Question

Giải phương trình

Answer 1

Ta có :

\(\left|x^2-x+2\right|-3x-7=0\)

\(\Leftrightarrow\left|\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right|=3x+7\)

Với mọi x ta có : \(\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\forall x\)

\(\Leftrightarrow\left|\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right|=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\)

\(\Leftrightarrow\left|x^2-x+2\right|=x^2-x+2\)

\(\Leftrightarrow x^2-x+2=3x+7\)

\(\Leftrightarrow x^2-4x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)

Vậy..