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B
24 tháng 9 2020
Bài làm :
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) Sửa đề : 5x3 + x2 - 4x + 9 = 0
<=>( 5x3 + 5 ) + (x2 - 4x +4)=0
<=> 5(x3 + 1) + (x-2)2 = 0
<=> 5(x+1)(x2 - x +1) + (x+2)2 =0
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}\\\end{cases}}\begin{cases}x=0\\x=-3\\x=2\end{cases}\)
24 tháng 9 2020
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) 5x3 + x2 - 4x - 9 = 0 ( đề sai )
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
<=> \(\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)( do 3x2 + 6 ≥ 6 > 0 với mọi x )
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
<=> x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = -3 hoặc x = 2
LP
0
GH
28 tháng 2 2018
a) \(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy ..................
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy .................
c) \(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+3\right)\left(2x+1+x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy .......................
d) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy ...................
TT
28 tháng 2 2018
a,\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...
b,\(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Vậy...
c,Sửa đề:
\(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3+2x+1\right)\left(x-3-2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\-x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-4\end{matrix}\right.\)
Vậy...
d,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+4=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-4\\x=3\end{matrix}\right.\)
Vậy...
22 tháng 1 2020
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
NH
22 tháng 1 2020
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
21 tháng 10 2020
a) \(7x-10=5x-6\)
\(7x-5x=-6+10\)
\(2x=4\)
\(x=2\)
b) \(3x\left(x-2\right)+x-2=0\)
\(\left(x-2\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
c) \(2x^2+7x-4=0\)
\(2x^2-x+8x-4=0\)
\(x\left(2x-1\right)+2\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
21 tháng 10 2020
7x-10=5x-6<=>7x-5x=-6+10<=>2x=4=>x=2
3x(x-2)+x-2=0<=>(x-2)(3x+1)=0<=>x-2=0=>x=2 HAY 3x+1=0=>x=-1/3
2x2+7x-4=0.
Câu cuối xem có lộn đề không nha bạn ơi!!!
6 tháng 7 2018
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
12 tháng 7 2015
\(1;x^2+7x+10=0\Rightarrow x^2+2x+5x+10=0\Rightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+5\right)=0\)
=> x + 2 = 0 hoặc x + 5 = 0
=> x = -2 hoặc x = - 5
2, x^4 - 5x^2 + 4 = 0
x^4 - 4x^2 - x^2 + 4 = 0
x^2 ( x^2 - 4) - ( x^2 - 4) = 0
( x^2 - 1)( x^2 - 4) = 0
( x - 1 )( x + 1)( x - 2)( x + 2) = 0
=> x= 1 hoặc x= -1 hoặc x = 2 hoặc x = - 2
Đúng cho mi8nhf mình giải tiếp cho
15 tháng 12 2020
a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)
b, \(x^3+x^2-9x-9=0\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-9\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=-1;\pm3\)
c, \(x^2-3x-10=0\Leftrightarrow x^2+2x-5x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=5;-2\)
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
x2 - 5x = 0
<=> x( x - 5 ) = 0
<=> x = 0 hoặc x - 5 = 0
<=> x = 0 hoặc x = 5
x2 - 7x + 10 = 0
<=> x2 - 2x - 5x + 10 = 0
<=> x( x - 2 ) - 5( x - 2 ) = 0
<=> ( x - 2 )( x - 5 ) = 0
<=> x - 2 = 0 hoặc x - 5 = 0
<=> x = 2 hoặc x = 5
\(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow x=0;5\)
\(x^2-7x+10=0\Leftrightarrow x^2-2x-5x+10=0\)
\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow x=5;2\)