Question

Giả phương trính chứa giá trị tuyệt đối

a) G = (x+1)² + | x +10 | - x² - 12 = 0

b) H = | 4 - x | + x² - (5 + x).x = 0

Answer 1

a)

\(\left(I\right)\left\{{}\begin{matrix}x\ge-10\\\left(x+1\right)^2+\left(x+10\right)-x^2-12=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-10\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-10\\x=\dfrac{1}{3}\end{matrix}\right.\) => \(x=\dfrac{1}{3}\)

\(\left(II\right)\left\{{}\begin{matrix}x< 10\\\left(x+1\right)^2+\left(x+10\right)-x^2-12=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< -10\\x=21\end{matrix}\right.\)Loại

Kết luận

x=1/3 nghiệm duy nhất

b)

\(H=\left|4-x\right|+x^2-\left(5+x\right)x=0\)

\(H=\left|x-4\right|-5x=0\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x=\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\) nghiệm x=2/3