Đề bài yêu cầu gì?

rút gọn

\(\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}-1}\div\frac{2}{x-1}+\frac{1}{\sqrt{x}+1}.\)

=\(\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)\div\frac{2}{\left(\sqrt{x}-1\right)\times\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}+1}\)

\(=\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\frac{2+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\times\left(\sqrt{x}+1\right)}\)

\(=\frac{1+x}{\sqrt{x}\times\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\frac{\left(1+x\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{1+x}{\sqrt{x}}\)

A>0 chứ ko phải x>0

a: \(M=\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

Dạ cảm ơn nhiều

a: ĐKXĐ: x>=0; x<>4

b: \(A=\left(\frac{\sqrt{x}}{x-4}-\frac{1}{\sqrt{x}+2}\right):\frac{\sqrt{x}-2}{x-4}\)

\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{1}{\sqrt{x}+2}\right)\cdot\left(\sqrt{x}+2\right)\)

\(=\frac{\sqrt{x}-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}+2\right)=\frac{2}{\sqrt{x}-2}\)

A<0

=>\(\frac{2}{\sqrt{x}-2}<0\)

=>\(\sqrt{x}-2<0\)

=>\(\sqrt{x}<2\)

=>0<=x<4

đề kiểu này hả:\(x+\sqrt{x}+1=0\)

ừ, trả lời hộ cái.

a: \(\frac{-1}{\sqrt{x}-3}<-1\)

=>\(-\frac{1}{\sqrt{x}-3}+1<0\)

=>\(\frac{-1+\sqrt{x}-3}{\sqrt{x}-3}<0\)

=>\(\frac{\sqrt{x}-4}{\sqrt{x}-3}<0\)

=>\(3<\sqrt{x}<4\)

=>9<x<16

b: Để \(\frac{-1}{\sqrt{x}-3}\) là số nguyên thì -1⋮\(\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\lbrace1;-1\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace4;2\right\rbrace\)

=>x∈{16;4}