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

nhầm đề \(x\sqrt{12}+\sqrt{18}=x\sqrt{18}+\sqrt{27}\)

ah quên nhầm đề đề là : tìm x : \(x\sqrt{2}+\sqrt{18}=x\sqrt{18}+\sqrt{27}\)

xin lỗi đề sai 

a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

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

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

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

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

c: \(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=18\)

=>2 căn x-2=18

=>x-2=81

=>x=83

a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

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

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

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

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

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

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

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

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

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

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)

Lời giải:

\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}\)

\(=\sqrt{2}.\sqrt{(2+\sqrt{2})(2-\sqrt{2})}=\sqrt{2}.\sqrt{2^2-2}=2\)

\(y=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{\frac{2}{3}(9\sqrt{2}-6\sqrt{3}+3\sqrt{5})}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{2}{3}\)

Do đó:

\(E=\frac{1+xy}{x+y}-\frac{1-xy}{x-y}=\frac{1+\frac{4}{3}}{2+\frac{2}{3}}-\frac{1-\frac{4}{3}}{2-\frac{2}{3}}=\frac{9}{8}\)

ĐKXĐ: `x>=0`

`a,sqrtx=21`

`=>x=21(TMĐK)`

KL...

`b,3\sqrtx=18`

`<=>sqrtx=6`

`=>x=36(TMĐK)`

KL...

`c,sqrtx <=5`

`=>x<=25` kết hợp với điều kiện có `0<=x<=25`

KL....

`d,3sqrt(2x)>9`

`<=>sqrt(2x)>3`

`=>2x>9`

`<=>x>9/2(TMĐK)`

KL...

a. \(\sqrt{x}=21\)

Vì x\(\ge\) 0 nên bình phương 2 vế ta được:

x = 21² \(\Leftrightarrow\) x = 441

Vậy x = 441

b \(3\sqrt{x}=18\) \(\Leftrightarrow\sqrt{x}=18:3\Leftrightarrow x=\sqrt{6}\) 

Vì \(x\ge0\) nên bình phương ta được:

x = 6² \(\Leftrightarrow\) x = 36

Vậy x = 36

c. \(\sqrt{x}hoặc=\sqrt{5}\) 

\(\sqrt{x}\le\sqrt{5}\) (đk x \(\le\) 0)

\(\Rightarrow x\le5\) 

Kết hợp với đk \(\Rightarrow0\le x\le5\)  

d. \(3\sqrt{2x}>9\) 

\(\Rightarrow\sqrt{2}>3\) 

\(\Rightarrow2x>9\) 

\(\Rightarrow x>\dfrac{9}{2}\) 

Kết hợp với điều kiện \(\Rightarrow x>\dfrac{9}{2}\)