a) \(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(A=\sqrt{16+8\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}\)

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

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

b) \(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(B=\sqrt{25+10\sqrt{2}+2}-\sqrt{16+8\sqrt{2}+2}\)

\(A=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)

\(A=5+\sqrt{2}-4-\sqrt{2}=1\)

\(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3+8\sqrt{3}+16}-\sqrt{3+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot4+4^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}+1^2}\)

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

\(=\left|\sqrt{3}+4\right|-\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}+4-\left(\sqrt{3}+1\right)\)

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

\(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=\sqrt{2+10\sqrt{2}+25}-\sqrt{2+8\sqrt{2}+16}\)

\(=\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot5+5^2}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot4+4^2}\)

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

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

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

\(=\sqrt{2}+5-\sqrt{2}-4=1\)

\(a\)

\(\sqrt{11}+\sqrt{19}\)

\(=\)\(\sqrt{11+19}\)

\(=\)\(\sqrt{30}\)

\(=\)\(5,47\)

\(\sqrt{47}\)

\(=6,85\)

\(5,47\)\(< \)\(6,85\)

\(=>\)\(\sqrt{11}+\sqrt{19}\)\(< \)\(\sqrt{47}\)

\(b\)

\(\sqrt{7}+\sqrt{26}+1\)

\(=\)\(\sqrt{7+26}+1\)

\(=\)\(\sqrt{33}+1\)

\(=\)\(5,74+1\)

\(=\)\(6,74\)

\(\sqrt{63}\)

\(=\)\(7,93\)

\(6,74\)\(< \)\(7,93\)

\(=>\)\(\sqrt{7}+\sqrt{26}+1\)\(< \)\(\sqrt{63}\)

Học tốt!!!

Bạn chép lại đề được không?

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