Bài làm:
Ta có: \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-2^2=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta có : \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)
a/ 3x^2 - 8x + 4 = 0
<=> (3x-2)(x-2)=0
<=> x=2/3 or x=2
b/ 4x^2 - 4x - 3 = 0
<=> (2x+1)(2x-3)=0
<=> x=-1/2 or x=3/2
<=> 4x^2 + 2x - 6x - 3 =0
<=> (4x^2 + 2x) - (6x + 3) =0
<=> 2x(2x + 1) - 3(2x + 1)=0
<=> (2x - 3)(2x + 1) = 0
<=> [ 2x - 3 = 0
.......[ 2x + 1 = 0
<=> [x = 3/2
.......[ x = -1/2
Bài làm :
Ta có :
\(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1,5\\x=-0,5\end{cases}}\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
4x2 - 4x - 3 = 0
<=> 4x2 + 2x - 6x - 3 = 0
<=> 2x( 2x + 1 ) - 3( 2x + 1 ) = 0
<=> ( 2x - 3 )( 2x + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-3=0&2x+1=0&\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Bài làm
\(4x^2-4x-3=0\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
