a) Ta có:
\(A=2x^2+\left|7x-1\right|-\left(5-x+2x^2\right)\)
\(=2x^2+7x-1-5+x-2x^2\)
\(=\left(7x+x\right)-1-5\)
\(=8x-6\)
b) Lại có: \(2x^2+\left|7x-1\right|-\left(5-x+2x^2\right)=2\)
\(\Rightarrow2x^2+\left|7x-1\right|-5+x-2x^2=2\)
\(\Rightarrow\left|7x-1\right|+x=7\)
\(\Rightarrow\left|7x-1\right|=7-x\)
+) TH1: \(7x-1\ge0\Rightarrow7x\ge1\Rightarrow x\ge\dfrac{1}{7}\)
Khi đó: \(7x-1=7-x\)
\(\Rightarrow7x+x=1+7\)
\(\Rightarrow8x=8\Rightarrow x=1\) (tm)
+) TH2: \(7x-1< 0\Rightarrow7x< 1\Rightarrow x< \dfrac{1}{7}\)
Lúc đó: \(-7x+1=7-x\)
\(\Rightarrow-7x+x=-1+7\)
\(\Rightarrow-6x=6\Rightarrow x=-1\) (tm)
Vậy \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\).
