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

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}

A>0 chứ ko phải x>0

a.ĐKXĐ;\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

b.P=\(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{x-4}\)

=\(\frac{3x-6\sqrt{x}}{x-4}=\frac{3\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)=\(\frac{3\sqrt{x}}{\sqrt{x}+2}\)

c.P=2\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}=2\Leftrightarrow3\sqrt{x}=2\sqrt{x}+\text{4}\)\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

Vậy x=16

thần đồng

Bạn ơi,bạn tra trên gu gồ đi,nhiều lắm

\(\sqrt{x}=4\)

\(\sqrt{x}=\sqrt{16}\)

\(x=16\)

\(\sqrt{x+1}=5\)

\(\sqrt{x+1}=\sqrt{25}\)

\(x+1=25\)

\(x=24\)