Question

[NT Linh]

help me ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

a) Ta có: \(A=3\sqrt{8}-\sqrt{50}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=3\cdot2\sqrt{2}-5\sqrt{2}-\left(\sqrt{2}-1\right)\)

\(=6\sqrt{2}-5\sqrt{2}-\sqrt{2}+1\)

\(=1\)

Ta có: \(B=\dfrac{3}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{9x^2}}\)

\(=\dfrac{3}{x-1}\cdot\dfrac{\left|x-1\right|}{3x}\)

\(=\dfrac{-\left(x-1\right)}{x-1}\cdot\dfrac{1}{x}\)(Vì 0<x<1)

\(=-1\cdot\dfrac{1}{x}=-\dfrac{1}{x}\)

b) Để \(B=\dfrac{-2}{\sqrt{x}}\) thì \(\dfrac{-1}{x}=\dfrac{-2}{\sqrt{x}}\)

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{2\sqrt{x}}{x}\)

Suy ra: \(2\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)

hay \(x=\dfrac{1}{4}\)(nhận)

Vậy: Để \(B=\dfrac{-2}{\sqrt{x}}\) thì \(x=\dfrac{1}{4}\)

Answer 2

[NT Linh]

thank bạn nha