\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}\)

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

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

b: Thay x=9 vào A, ta được:

\(A=\dfrac{2\cdot\sqrt{9}}{9+1}=\dfrac{2\cdot3}{10}=\dfrac{6}{10}=\dfrac{3}{5}\)

c: Để A=1 thì \(\dfrac{2\sqrt{x}}{x+1}=1\)

=>\(x+1=2\sqrt{x}\)

=>\(x-2\sqrt{x}+1=0\)

=>\(\left(\sqrt{x}-1\right)^2=0\)

=>\(\sqrt{x}-1=0\)

=>\(\sqrt{x}=1\)

=>x=1(loại)

a,  \(A=\frac{x^2+3x-x+3-x^2+1}{x^2-9}\)\(.\frac{x+3}{2}\)            \(\left(x\ne3;-3\right)\)

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

b, để \(A\in Z\Rightarrow\hept{\begin{cases}x+2⋮x-3\\x-3⋮x-3\end{cases}}\)\(\Rightarrow x+2-x+3=5⋮x-3\)\(\leftrightarrow x+3\in\left(1;5;-1;-5\right)\)

                                                                                                                              \(\leftrightarrow x\in\left(-2;2;-4;-8\right)\)

Mới 2k9

2k9 thì thôi :)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;1\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{x-4\sqrt{x}+3-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2x-4\sqrt{x}+5-2x+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)

\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)

b: Để A>2 thì A-2>0

=>\(\dfrac{1-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)

=>\(\dfrac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)

=>\(\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}< 0\)

TH1: \(\left\{{}\begin{matrix}2\sqrt{x}-5>0\\\sqrt{x}-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>\dfrac{5}{2}\\\sqrt{x}< 2\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}2\sqrt{x}-5< 0\\\sqrt{x}-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< \dfrac{5}{2}\\\sqrt{x}>2\end{matrix}\right.\)

=>\(2< \sqrt{x}< \dfrac{5}{2}\)

=>4<x<25/4

c: Để A là số nguyên thì \(1⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)

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

=>\(x\in\left\{1;9\right\}\)

kết hợp ĐKXĐ, ta được: x=9

a) Ta có: \(A=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)

\(=\dfrac{2x^2+1}{x^2+1}:\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\dfrac{2x^2+1}{x^2+1}\cdot\dfrac{\left(x-1\right)\left(x^2+1\right)}{\left(x-1\right)^2}\)

\(=\dfrac{2x^2+1}{x-1}\)

b) Thay \(x=-\dfrac{1}{2}\) vào A, ta được:

\(A=\left(2\cdot\dfrac{1}{4}+1\right):\left(\dfrac{-1}{2}-1\right)\)

\(=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)

c) Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{2x^2+1}{x-1}-1< 0\)

\(\Leftrightarrow\dfrac{2x^2+1-x+1}{x-1}< 0\)

\(\Leftrightarrow\dfrac{2x^2-x+2}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\)

hay x<1