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3 tháng 8 2019
1) \(x^3-x^2=4x^2-8x+4\)
\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^2-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-2x.2+2^2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
17 tháng 10 2016
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
b) \(\left(2x-3\right)^2=\left(x-5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-3+x-5\right)\left(2x-3-x+5\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{8}{3}\end{array}\right.\)
c) \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
27 tháng 6 2017
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
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
<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)
27 tháng 10 2020
Bài 3:
a) Ta có: \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4;-4\right\}\)
b) Ta có: \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)
c) Ta có: \(\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\)
d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
NT
0
20 tháng 11 2017
Bài 1:
a) \(\left(8x^4-4x^3+x^2\right):2x^2\)
\(=4x^2-2x+1\)
b) \(\left(6x^4-3x^3+x^2\right):3x^2\)
\(=2x^2-x+3\)
Bài 2:
\(2x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\2x-1=0\Rightarrow x=0,5\end{matrix}\right.\)