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6 tháng 8 2017
a,4x^2-4x+1=0
4x^2-2x-2x+1=0
2x (2x-1)-(2x-1)=0
(2x-1)(2x-1)=0
(2x-1)^2=0
=>2x-1=0 <=> x=1/2
BT
5
2 tháng 10 2016
a, 4x^2 - 4x = -1
\(\Leftrightarrow\)4x^2 - 4x + 1 = 0
\(\Leftrightarrow\)(2x-1)2 =0
\(\Leftrightarrow\)2x - 1 = 0
\(\Leftrightarrow\)x = 1/2
b, \(\Leftrightarrow\)( 2x + 1)^3 = 0
\(\Leftrightarrow\)2x + 1 = 0
\(\Leftrightarrow\)x = -1/2
đúng thì
MA
2 tháng 10 2016
a) \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(8x^3+12x^2+6x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
28 tháng 8 2017
*) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
*)\(8x^3+12x^2+6x+1=0\Leftrightarrow8x^3+8x^2+2x+4x^2+4x+1\)
\(\Leftrightarrow2x\left(4x^2+4x+1\right)+\left(x^2+4x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1\right)^2=0\Leftrightarrow\left(2x+1\right)^3=0\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
28 tháng 8 2017
a,42 - 4x = -1
=> 4x2 - 4x + 1 = 0
=> (2x - 1)2 = 0
=> 2x -1 = 0
=> x = \(\dfrac{1}{2}\)
b, 8x3 +12x2 + 6x +1=0
=> ( 2x +1 )3 = 0
=> 2x + 1 = 0
=> x = \(\dfrac{-1}{2}\)
MO
5
M
6 tháng 8 2019
a) \(x^2-12x+11\)\(=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6+5\right)\left(x-6-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
6 tháng 8 2019
a)\(x^2-12x+11=0\)
\(x^2-x-11x+11=0\)
\(\left(x^2-x\right)-\left(11x-11\right)=0\)
\(x\left(x-1\right)-11\left(x-1\right)=0\)
\(\left(x-1\right)\left(x-11\right)=0\)
\(=>\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
b)\(4x^2-4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(3x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(=>\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0,5\\x=-1,5\end{matrix}\right.\)\
c)\(4x^2-12x-7=0\)
\(4x^2-14x+2x-7=0\)
\(2x\left(2x-7\right)+\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(2x+1\right)=0\)
\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
11 tháng 10 2020
a) \(x\left(x-2\right)-7x+14=0\)
\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(x^2+12x-13=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(13x-13\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+13\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) \(4x^2-4x=8\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) \(x^2-6x=1\)
\(\Leftrightarrow\left(x-3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
11 tháng 10 2020
a) x( x - 2 ) - 7x + 14 = 0
<=> x( x - 2 ) - 7( x - 2 ) = 0
<=> ( x - 2 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) x2( x - 3 ) + 12 - 4x = 0
<=> x2( x - 3 ) - 4( x - 3 ) = 0
<=> ( x - 3 )( x2 - 4 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) x2 + 12x - 13 = 0
<=> x2 - x + 13x - 13 = 0
<=> x( x - 1 ) + 13( x - 1 ) = 0
<=> ( x - 1 )( x + 13 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) 4x2 - 4x = 8
<=> 4( x2 - x ) = 8
<=> x2 - x = 2
<=> x2 - x - 2 = 0
<=> x2 + x - 2x - 2 = 0
<=> x( x + 1 ) - 2( x + 1 ) = 0
<=> ( x + 1 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) x2 - 6x = 1
<=> x2 - 6x + 9 = 1 + 9
<=> ( x - 3 )2 = 10
<=> ( x - 3 )2 = ( ±√10 )2
<=> \(\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)