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HH
25 tháng 6 2018
\(x^3+9x=0\)
<=> \(x\left(x^2+9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)
<=> \(x=0\)
\(9x^2-4-2\left(3x-2\right)^2=0\)
<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)
<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)
<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)
<=> \(3\left(3x-2\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
\(\left(x^3-x^2\right)-4x+8x-4=0\)
<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+4\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)
<=> \(x=1\)
\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)
<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)
<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)
<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)
<=> \(5x-2+3x-6=-4\)
<=> \(8x-8=-4\)
<=> \(8\left(x-1\right)=-4\)
<=> \(x-1=-\frac{1}{2}\)
<=> \(x=-\frac{3}{2}\)
21 tháng 10 2019
a/ \(\left(2x-3\right)^2-\left(3x+2\right)^2=5x\left(2-x\right)\)
<=> \(\left(2x-3-3x-2\right)\left(2x-3+3x+2\right)=5x\left(2-x\right)\)
<=> \(\left(-x-5\right)\left(5x-1\right)=5x\left(2-x\right)\)
<=> \(-5x^2-25x+x+5=10x-5x^2\)
<=> \(10x+25x-x=5\)
<=> \(34x=5\)
<=> \(x=\frac{5}{34}\)
b/ pt <=> \(2^3x^3-3.2^2.x^2.1+3.2.x.1^2-1^3=0\)
<=> \(\left(2x-1\right)^3=0\)
<=> 2 x - 1 = 0
<=> x = 1/2.
S
13 tháng 10 2018
\(a.x^4-16x^2=0\Leftrightarrow\left(x^2+4x\right)\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
\(b.\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
NL
13 tháng 10 2018
a) x4 - 16x2 = 0
<=> x2 ( x2 - 16 ) = 0
<=> \(\left[{}\begin{matrix}x^2=0\\x^2-16=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
Vậy...
b) ( x - 5)3 - x + 5 = 0
<=> ( x - 5)3 - (x - 5) = 0
<=> (x - 5) [ (x - 5)2 - 1] =0
<=> \(\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy...
c) 5(x - 2) = x2 - 4
<=> 5(x - 2) - (x2 - 4) = 0
<=> (x - 2)( 5 - x - 2) = 0
<=> (x - 2)( 3 - x ) = 0
<=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy...
d) x - 3 = (3 - x)2
<=> x - 3 - (x - 3)2 = 0
<=> (x - 3)(1 - x + 3) = 0
<=> (x - 3)( 4 - x ) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy...
e) x2 (x - 5) + 5 - x = 0
<=> x2 (x - 5) - (x - 5) = 0
<=> (x2 - 1)( x - 5) = 0
<=> \(\left[{}\begin{matrix}\left(x-1\right)\left(x+1\right)=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
,
XO
1 tháng 10 2020
a) (2x - 3)2 = (x + 5)2
=> 4x2 - 12x + 9 = x2 + 10x + 25
=> 4x2 - 12x + 9 - (x2 + 10x + 25) = 0
=> 3x2 - 22x - 16 = 0
=> 3x2 - 24x + 2x - 16 = 0
=> 3x(x - 8) + 2(x - 8) = 0
=> (3x + 2)(x - 8) = 0
=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)
b) x2(x - 1) - 4x2 + 8x - 4 = 0
=> x2(x - 1) - (2x - 2)2 = 0
=> x2(x - 1) - [2(x- 1)]2 = 0
=> x2(x - 1) - 4(x - 1)2 = 0
=> (x - 1)(x2 - 4(x - 1) = 0
=> (x - 1)(x2 - 4x + 4) = 0
=> (x - 1)(x - 2)2 = 0
=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
c) x2 + 7x + 12 = 0
=> x2 + 3x + 4x + 12 = 0
=> x(x + 3) + 4(x + 3) = 0
=> (x + 4)(x + 3) = 0
=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)
d) x2 + 3x - 18 = 0
=> x2 + 6x - 3x - 18 = 0
=> x(x + 6) - 3(x + 6) = 0
=> (x - 3)(x + 6) = 0
=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
e) x(x + 6) - 7x - 42 = 0
=> x(x + 6) - 7(x + 6) = 0
=> (x - 7)(x + 6) = 0
=> \(\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
1 tháng 10 2020
1. ( 2x - 3 )2 = ( x + 5 )2
<=> ( 2x - 3 )2 - ( x + 5 )2 = 0
<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0
<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0
<=> ( x - 8 )( 3x + 2 ) = 0
<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
2. x2( x - 1 ) - 4x2 + 8x - 4 = 0
<=> x2( x - 1 ) - ( 4x2 - 8x + 4 ) = 0
<=> x2( x - 1 ) - 4( x2 - 2x + 1 ) = 0
<=> x2( x - 1 ) - 4( x - 1 )2 = 0
<=> ( x - 1 )[ x2 - 4( x - 1 ) ] = 0
<=> ( x - 1 )( x2 - 4x + 4 ) = 0
<=> ( x - 1 )( x - 2 )2 = 0
<=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
3. x2 + 7x + 12 = 0
<=> x2 + 3x + 4x + 12 = 0
<=> x( x + 3 ) + 4( x + 3 ) = 0
<=> ( x + 3 )( x + 4 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
4. x2 + 3x - 18 = 0
<=> x2 - 3x + 6x - 18 = 0
<=> x( x - 3 ) + 6( x - 3 ) = 0
<=> ( x - 3 )( x + 6 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
5. x( x + 6 ) - 7x - 42 = 0
<=> x( x + 6 ) - 7( x + 6 ) = 0
<=> ( x + 6 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)
GN
7
N
22 tháng 10 2018
\(x^3-11x^2+30x=0\)
\(\left(x-6\right).\left(x-5\right).x=0\)
\(=>\orbr{\begin{cases}x-6=0\\x-5=0,x=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=5,x=0\end{cases}}\)
P/S: mk mới lớp 7 sai sót mong bỏ qua
N
22 tháng 10 2018
\(8x^2+30x+7=0\)
\(8x^2+28x+2x+7=0\)
\(2x.\left(4x+1\right)+7.\left(4x+1\right)=0\)
\(\left(2x+7\right).\left(4x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=-7\\4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
vậy ....
P/S sorry mk làm hơi lâu :)__chờ tí làm câu a cho
9 tháng 2 2017
Đặt 8x^2=t => t>=0
\(t^2-9-\left(t-1\right)=54\Leftrightarrow t^2-t+1-9=54\)
\(t^2-t+\frac{1}{4}=54+8+\frac{1}{4}=\frac{249}{4}\) lẻ thế nhỉ
\(\left(t-\frac{1}{2}\right)^2=\frac{249}{4}\Rightarrow\left[\begin{matrix}t=\frac{1-\sqrt{249}}{2}< 0\left(loai\right)\\t=\frac{1+\sqrt{249}}{2}\end{matrix}\right.\)
\(\left\{\begin{matrix}x< 0\\8x^2=\frac{1+\sqrt{249}}{2}\end{matrix}\right.\Rightarrow x=\frac{-\sqrt{1+\sqrt{249}}}{16}\)