Question

Bài 4: a) Rút gọn Q. Q=2x- căn(x^2-2x+1) . b) tìm x để Q=7

Answer 1

a) ĐKXĐ: \(x^2-2x+1\ge0\Rightarrow\left(x-1\right)^2\ge0\left(luônđúng\right)\)

\(Q=2x-\sqrt{x^2-2x+1}=2x-\sqrt{\left(x-1\right)^2}=2x-x+1=x+1\)

b) \(Q=x+1\\ \Rightarrow7=x+1\\ \Rightarrow x=6\)

Answer 2

a: Ta có: \(Q=2x-\sqrt{x^2-2x+1}\)

\(=2x-\left|x-1\right|\)

\(=\left[{}\begin{matrix}2x-x+1=x+1\left(x\ge1\right)\\2x+x-1=3x-1\left(x< 1\right)\end{matrix}\right.\)

b: Ta có: Q=7

nên \(\left[{}\begin{matrix}x+1=7\left(x\ge1\right)\\3x-1=7\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{8}{3}\left(loại\right)\end{matrix}\right.\)