Ta có : \(ad=bc;a,b,c,d>0\)

\(\Rightarrow2\sqrt{ad}=2\sqrt{bc}\)

Khi đó : \(\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\) \(=\frac{1}{\left(\sqrt{a}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{d}\right)-\left(\sqrt{b}+\sqrt{c}\right)}{\left[\left(\sqrt{a}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)\right].\left[\left(\sqrt{a}+\sqrt{d}\right)-\left(\sqrt{b}+\sqrt{c}\right)\right]}\)

\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{\left(\sqrt{a}+\sqrt{d}\right)^2-\left(\sqrt{b}+\sqrt{c}\right)^2}\) \(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+d+2\sqrt{ad}-b-c-2\sqrt{bc}}\)

\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+d-b-c}\) ( Do \(2\sqrt{ad}=2\sqrt{bc}\) )

\(\frac{x}{4}=\frac{y}{9}\Rightarrow9x=4y\)      

\(y=\frac{9}{4}x\)

P=\(\frac{-\sqrt{x}}{\sqrt{x}+\sqrt{\frac{9}{4}x}}\)             

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

\(=\frac{-\sqrt{x}}{\frac{5}{2}\sqrt{x}}\)   ( ĐK : x > 0 ) 

\(=-\frac{2}{5}\)      

a, \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=140\Leftrightarrow x=28\)

b,ĐK :  \(x\ge7\)

 \(\sqrt{x^2-14x+49}-3x=1\Leftrightarrow\sqrt{\left(x-7\right)^2}=3x+1\)

\(\Leftrightarrow x-7=3x+1\Leftrightarrow-2x-8=0\Leftrightarrow x=-4\)( vô lí )

c, Bn làm nốt nhé 

a) đk: \(x\le\frac{4}{5}\)

Ta có: \(\sqrt{4-5x}=12\)

\(\Leftrightarrow\left|4-5x\right|=144\)

\(\Rightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

\(\Rightarrow x=-28\left(tm\right)\)

b) Ta có: \(\sqrt{x^2-14x+49}-3x=1\)

\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=1+3x\)

\(\Leftrightarrow\left|x-7\right|=3x+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=3x+1\\x-7=-3x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-8\\4x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)

a) Ta có: \(\sqrt{16-6\sqrt{7}}+\sqrt{7}\)

\(=\sqrt{3^2-2.3.\sqrt{7}+7}+\sqrt{7}\)

\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}\)

\(=\left|3-\sqrt{7}\right|+\sqrt{7}\)

\(=3-\sqrt{7}+\sqrt{7}\)

\(=3\)

b) Ta có: \(\sqrt{\left|12\sqrt{5}-29\right|}+\sqrt{12\sqrt{5}+29}\)

\(=\sqrt{\left(\sqrt{29-12\sqrt{5}}+\sqrt{12\sqrt{5}+29}\right)^2}\)

\(=\sqrt{29-12\sqrt{5}+2\sqrt{\left(29-12\sqrt{5}\right)\left(12\sqrt{5}+29\right)}+12\sqrt{5}+29}\)

\(=\sqrt{58+2\sqrt{121}}\)

\(=\sqrt{58+2.11}\)

\(=\sqrt{80}=4\sqrt{5}\)

Bài làm:

Đặt \(A=\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}\)

\(\Leftrightarrow A^2=\left(\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}\right)^2\)

\(=7-\sqrt{13}-2\sqrt{\left(7-\sqrt{13}\right)\left(7+\sqrt{13}\right)}+7+\sqrt{13}\)

\(=14-2\sqrt{49-13}\)

\(=14-2\sqrt{36}=14-2.6=14-12=2\)

\(\Rightarrow A=\sqrt{2}\)

Thay vào ta được:

\(\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}+\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

thanks bạn nha

Bài làm:

Ta có: 

\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)

\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)

\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)

a) \(a>\sqrt{a}\Leftrightarrow a^2>a\)   

\(a^2-a>0\) 

\(a\left(a-1\right)>0\)   

\(\hept{\begin{cases}a>0\\a-1>0\end{cases}}\) 

\(\hept{\begin{cases}a>0\\a>1\end{cases}}\)    

\(\Rightarrow a>1\)           

b) 

\(a< \sqrt{a}\)    

\(a^2< a\)       

\(a^2-a< 0\)   

\(a\left(a-1\right)< 0\) 

\(\hept{\begin{cases}a>0\\a-1< 0\end{cases}}\)   

\(\hept{\begin{cases}a>0\\a< 1\end{cases}}\)