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

\(=\left(\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

\(=2\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{2a-4}{a+2}\)

Ta có: \(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

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

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

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

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

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

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

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

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

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

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

\(B=\dfrac{2\sqrt{x}-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)

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

\(a,A=2\sqrt{3}+10\sqrt{3}-5\sqrt{3}=7\sqrt{3}\\ b,ĐK:x\ge0\\ PT\Leftrightarrow x^2-2x+1+x-2\sqrt{x}+1=0\\ \Leftrightarrow\left(x-1\right)^2+\left(\sqrt{x}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow x=1\left(tm\right)\)

\(P=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)ĐK : x > 0 

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

\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

bạn bổ sung đk hộ mình ý 2 là : \(x\ge0;x\ne1\)nhé 

1. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}=13-x$

\(\Rightarrow \left\{\begin{matrix} 13-x\geq 0\\ x-1=(13-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 13\\ x^2-27x+170=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 13\\ (x-17)(x-10)=0\end{matrix}\right.\)

\(\Rightarrow x=10\) (tm)

2. ĐKXĐ: $x\geq 3$

\(3\sqrt{x+34}-3\sqrt{x-3}=1\)

\(\Leftrightarrow 3\sqrt{x+34}=3\sqrt{x-3}+1\)

\(\Rightarrow 9(x+34)=9x+6\sqrt{x-3}-26\)

\(\Leftrightarrow \frac{166}{3}=\sqrt{x-3}\)

$\Leftrightarrow x-3=\frac{27556}{9}$

$\Leftrightarrow x=\frac{27583}{9}$ (tm)

Bài 7:

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

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

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