1. \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)

\(\Leftrightarrow\left(\sqrt{10+\sqrt{3x}}\right)^2=\left(2+\sqrt{6}\right)^2\)

\(\Leftrightarrow10+\sqrt{3x}=4+6\)

\(\Leftrightarrow10+\sqrt{3x}=10\)

\(\Leftrightarrow\sqrt{3x}=0\)

\(\Leftrightarrow x=0\)

1)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)

\(\Leftrightarrow3,185586233=2+\sqrt{6}\)

Tương tự làm tiếp bài này đi! Mình lười quá!

2) Tương tự!

2.\(\sqrt{4x+20}-3\sqrt{5+x}+\frac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{5+x}+\frac{4}{3}\sqrt{9\left(x+5\right)}=6\)

\(\Leftrightarrow\sqrt{x+5}\left(\sqrt{4}-3+\frac{4}{3}\right)=6\)

\(\Leftrightarrow\sqrt{x+5}\left(2-3+\frac{3}{4}\right)=6\)

\(\Leftrightarrow\sqrt{x+5}\left(\frac{-1}{4}\right)=6\)

\(\Leftrightarrow\sqrt{x+5}=6:\left(-\frac{1}{4}\right)=-24\)

\(\Leftrightarrow\)K tồn tại x

a: \(P=\dfrac{15\sqrt{x}-11+\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11+3x+7\sqrt{x}-6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+21\sqrt{x}-14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

b: Khi x=9 thì \(P=\dfrac{9+21\cdot3-14}{\left(3+3\right)\left(3-1\right)}=\dfrac{29}{6}\)

Ta có: \(A=\frac{1}{3}+\frac{1}{6}+......+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)

       \(A=\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{2002}.\frac{1}{2}\)

   \(A=\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{4004}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2000}{4004}\)

\(A=\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{4004}\)

\(A=\frac{1}{x+1}=\frac{1}{2}-\frac{2000}{4004}\)

\(A=\frac{1}{x+1}=\frac{1}{2002}\)

\(x+1=2002\)

nên \(x=2002-1=2001\)

Vậy x = 2001

tự mà lm để IQlên chứ

ĐKXĐ : \(x\ge3\)

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

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

\(\Leftrightarrow\sqrt{x-3}+1+\sqrt{x-3}+3=4\)

\(\Leftrightarrow2\sqrt{x-3}=0\Leftrightarrow x=3\)(TMĐK)

\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)

a) \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,9\right)\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

b) \(\sqrt{x}=\sqrt{6+4\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)

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

\(=4\sqrt{2}+5\)

c) \(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Để \(Q\in Z\Rightarrow4⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{4;5;7;2;1\right\}\Rightarrow x\in\left\{16;25;49;4;1\right\}\)

a) Ta có: \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

b) Thay \(x=6+4\sqrt{2}\) vào Q, ta được:

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

a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)

b: Ta có: P=A:B

\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)

\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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