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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
KT
16 tháng 7 2018
a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
XO
21 tháng 12 2020
a) x(x - 3) + 5x = x2 - 8
=> x2 - 3x + 5x - x2 + 8 = 0
=> 2x + 8 = 0
=> 2x = -8
=> x = -4
b) 3(x + 4) - x2 - 4x = 0
=> 3(x + 4) - x(x + 4) = 0
=> (3 - x)(x + 4) = 0
=> \(\orbr{\begin{cases}3-x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
Vậy \(x\in\left\{3;-4\right\}\)là giá trị cần tìm
c) 7x3 + 12x2 - 4x = 0
=> x(7x2 + 12x - 4) = 0
=> x(7x2 + 14x - 2x - 4) = 0
=> x[7x(x + 2) - 2(x + 2)] = 0
=> x(x + 2)(7x - 2) = 0
=> x = 0 hoặc x + 2 = 0 hoặc 7x - 2 = 0
=> x = 0 hoặc x = -2 hoặc x = 2/7
Vậy \(x\in\left\{0;-2;\frac{2}{7}\right\}\)là giá trị cần tìm
21 tháng 12 2020
x( x - 3 ) + 5x = x2 - 8
⇔ x2 - 3x + 5x - x2 + 8 = 0
⇔ 2x + 8 = 0
⇔ 2x = -8
⇔ x = -4
3( x + 4 ) - x2 - 4x = 0
⇔ 3( x + 4 ) - x( x + 4 ) = 0
⇔ ( x + 4 )( 3 - x ) = 0
⇔ x = -4 hoặc x = 3
7x3 + 12x2 - 4x = 0
⇔ x( 7x2 + 12x - 4 ) = 0
⇔ x( 7x2 + 14x2 - 2x - 4 ) = 0
⇔ x[ 7x( x + 2 ) - 2( x + 2 ) ] = 0
⇔ x( x + 2 )( 7x - 2 ) = 0
⇔ x = 0 hoặc x = -2 hoặc x= 2/7
S
3
22 tháng 10 2020
1) ( x - 1 )3 - ( x + 3 )( x2 - 3x + 9 ) + 3( x2 - 4 ) = 2
⇔ x3 - 3x2 + 3x - 1 - ( x3 + 27 ) + 3x2 - 12 = 2
⇔ x3 + 3x - 13 - x3 - 27 = 2
⇔ 3x - 40 = 2
⇔ 3x = 42
⇔ x = 14
2) ( x2 - 4x )2 - 8( x2 - 4x ) + 15 = 0
Đặt t = x2 - 4x
pt ⇔ t2 - 8t + 15 = 0
⇔ t2 - 3t - 5t + 15 = 0
⇔ t( t - 3 ) - 5( t - 3 ) = 0
⇔ ( t - 3 )( t - 5 ) = 0
⇔ ( x2 - 4x - 3 )( x2 - 4x - 5 ) = 0
⇔ \(\orbr{\begin{cases}x^2-4x-3=0\\x^2-4x-5=0\end{cases}}\)
+) x2 - 4x - 3 = 0
⇔ ( x2 - 4x + 4 ) - 7 = 0
⇔ ( x - 2 )2 - ( √7 )2 = 0
⇔ ( x - 2 - √7 )( x - 2 + √7 ) = 0
⇔ \(\orbr{\begin{cases}x-2-\sqrt{7}=0\\x-2+\sqrt{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{7}\\x=2-\sqrt{7}\end{cases}}\)
+) x2 - 4x - 5 = 0
⇔ x2 - 5x + x - 5 = 0
⇔ x( x - 5 ) + ( x - 5 ) = 0
⇔ ( x - 5 )( x + 1 ) = 0
⇔ x = 5 hoặc x = -1
Vậy ...
22 tháng 10 2020
Bài làm
(x - 1)3 - (x + 3)(x2 - 3x + 9) + 3(x2 - 4) = 2
<=> x3 - 3x2 + 3x - 1 - (x3 + 33) + 3x2 - 12 = 2
<=> x3 - 3x2 + 3x - 1 - x3 - 27 + 3x2 - 12 - 2 = 0
<=> 3x - 42 = 0
<=> 3x = 42
<=> x = 14
Vậy nghiệm của phương trình là 4.
(x2 - 4x)2 - 8(x2 - 4x) + 15 = 0
Đặt x2 - 4x = t, ta có:
t2 - 8t + 15 = 0
<=> t2 - 3t - 5t + 15 = 0
<=> t(t - 3) - 5(t - 3) = 0
<=> (t - 5)(t - 3) = 0
<=> \(\orbr{\begin{cases}t-5=0\\t-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=5\\t=3\end{cases}}\)
Thay: t = 5 vào x2 - 4x ta được:
x2 - 4x = 5
<=> x2 - 4x - 5 = 0
<=> x2 - 5x + x - 5 = 0
<=> x(x - 5) + (x - 5) = 0
<=> (x + 1)(x - 5) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}}\)
Thay: t = 3 vào x2 - 4x ta được:
x2 - 4x = 3
<=> x2 - 4x - 3 = 0
<=> x2 - 4x + 4 - 7 = 0
<=> (x - 2)2 - 7 = 0
<=> (x - 2)2 = V 7
<=> x - 2 = + V 7
<=> \(\orbr{\begin{cases}x-2=-7\\x-2=7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\sqrt{7}+2\\x=\sqrt{7}+2\end{cases}}}\)
Vậy x = { -1; 5; \(-\sqrt{7}+2;\sqrt{7}+2\)}
LD
18 tháng 9 2018
Bài 1:
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(114x^2+216x+81=114x^2-480x+400\)
\(144x^2+216x=144x^2-480x+400-81\)
\(114x^2+216=114x^2-480x+319\)
\(696x=319\)
\(\Rightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Rightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Bài 2:
a) \(5x^3-7x^2-15x+21=0\)
\(\left(5x-7\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Rightarrow x=\frac{7}{5}\)
b) \(\left(x-3\right)^2=4x^2-20x+25\)
\(x^2-6x+9-25=4x^2-20x+25\)
\(x^2-6x+9=4x^2-20x+25-25\)
\(x^2-6x-16=4x^2-20x\)
\(x^2+14x-16=4x^2-4x^2\)
\(-3x^2+14x-16=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
c) \(\left(x-1\right)^2-5=\left(x+2\right)\left(x-2\right)-x\left(x-1\right)\)
\(x^2-2x=x-4\)
\(x^2-2x=x-4+4\)
\(x^2-2x=x-x\)
\(x^2-3x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
d) \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)
\(-48x^2+56x-24=-24\)
\(-48x^2+56x=-24+24\)
\(-48x^2+56=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{6}\end{cases}}\)
mình ko chắc
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: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
PT
3
LP
1
ML
16 tháng 8 2015
a)x2-20-x=0
<=>(x2-5x)+(4x-20)=0
<=>x(x-5)+4(x-5)=0
<=>(x-5)(x+4)=0
<=>x-5=0 hoặc x+4=0
<=>x=5 hoặc x=-4
b)(2x+3)2-(4x2-9)=0
<=>(2x+3)(2x+3)-(2x-3)(2x+3)=0
<=>(2x+3)(2x+3-2x+3)=0
<=>(2x+3).6=0
<=>2x+3=0
<=>2x=-3
<=>x=-1,5
c)(2x2+5x+3):(x+1)=4x-5
<=>2x2+5x+3=(4x-5)(x+1)
<=>2x2+5x+3=4x2-x-5
<=>4x2-x-5-2x2-5x-3=0
<=>2x2-6x-8=0
<=>x2-3x-4=0
<=>(x2-4x)+(x-4)=0
<=>x(x-4)+(x-4)=0
<=>(x-4)(x+1)=0
<=>x+1=0 hoặc x-4=0
<=>x=-1 hoặc x=4
DL
4
HT
19 tháng 12 2016
a, 5x - 7(3 - x) = 3
=> 5x - 21 + 7x = 3
=> 12x = 24
=> x = 2
b, 4x2 + 3x = 0
=> x(4x + 3) = 0
=> \(\orbr{\begin{cases}x=0\\4x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{-3}{4}\end{cases}}\)
c, (x + 1)2 - 4x2 =0
=> (x + 1)2 - (2x)2 = 0
=> (x + 1 - 2x)(x + 1 + 2x) = 0
=> (1 - x)(3x+ 1) = 0
=> \(\orbr{\begin{cases}1-x=0\\3x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
d, x3 - 19x - 30 = 0
=> x3 - 5x2 + 5x2 - 25x + 6x - 30 = 0
=> x2(x - 5) + 5x(x - 5) + 6(x - 5) = 0
=> (x2 + 5x + 6)(x - 5) = 0
=> (x2 + 2x + 3x + 6)(x - 5) = 0
=> (x + 2)(x + 3)(x - 5) = 0
=> x + 2 = 0 hoặc x + 3 = 0 hoặc x - 5 = 0
=> x = -2 hoặc x = -3 hoặc x = 5
=> x thuộc {-2; -3; 5}
\(4x^2-4x\left(x-3\right)+\left(3-x\right)^2=0\)
\(\Leftrightarrow4x^2-4x\left(x-3\right)+\left(x-3\right)^2=\left[\left(2x\right)-\left(x-3\right)\right]^2=\left(x+3\right)^2=0\)
\(x+3=0\Leftrightarrow x=-3\)