Môn Toán Lớp 8 Tìm x 4x^2 – 1 – (2x +1) = 0
Question
Kennedy
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Trả lời ( )
Ta có :4x^2 – 1 – (2x +1) = 0
=)( 2x-1)(2x+1)-(2x+1)=0
=)(2x+1)(2x-1-1)=0
=)(2x+1)(2x-2)=0
=)2x+1=0hoặc 2x-2=0
=)x=-1/2 hoặc x=1
$4x^2-1-(2x+1)=0_{}$
$⇔4x^2-1-2x-1=0_{}$
$⇔4x^2-2x-2=0_{}$
$⇔2.(2x^2-x-1)=0_{}$
$⇔2.(2x^2-2x+x-1)=0_{}$
$⇔2.[ (2x^2-2x)+(x-1)]=0_{}$
$⇔2.[ 2x.(x-1)+x.(x-1)]=0_{}$
$⇔2.(2x+1)(x-1)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}2x+1=0\\x-1=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-\frac{1}{2}\\x=1\end{array} \right.\)
$Vậy_{}$ $x_{1}=$ $-\frac{1}{2};$ $x_{2}=1$