14, \(\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)

\(=\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}\)

\(=\frac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(3\sqrt{x}+14\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+14}{5\sqrt{x}-1}\)

thank

\(\frac{3\sqrt{2}-6}{\sqrt{2}-1}=\frac{3\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=3\)

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

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

2) \(\frac{1}{\sqrt{5}+2}+\frac{\sqrt{15}+\sqrt{10}}{\sqrt{2}+\sqrt{5}}=\frac{\sqrt{2}+\sqrt{5}+\left(\sqrt{15}+\sqrt{10}\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{2}+\sqrt{5}\right)}\)

\(=\frac{\sqrt{2}+\sqrt{5}+5\sqrt{3}+2\sqrt{15}+5\sqrt{2}+2\sqrt{10}}{\left(\sqrt{5}+2\right)\left(\sqrt{2}+\sqrt{5}\right)}\)

...

a) \(A=\frac{3}{\sqrt{7}-2}+\frac{7}{\sqrt{7}-\sqrt{28}}=\frac{3}{\sqrt{7}-2}-\sqrt{7}\)

\(=\frac{3-7+2\sqrt{7}}{\sqrt{7}-2}=\frac{-4+2\sqrt{7}}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)

đk: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

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

\(B=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(B=\frac{2}{\sqrt{x}+2}\)

b) \(6B-A>0\Leftrightarrow\frac{12}{\sqrt{x}+2}-2>0\)

\(\Leftrightarrow\frac{8-2\sqrt{x}}{\sqrt{x}+2}>0\Rightarrow8-2\sqrt{x}>0\left(because:\sqrt{x}+2>0\right)\)

\(\Rightarrow\sqrt{x}< 4\Rightarrow x< 16\)

Vậy \(\hept{\begin{cases}0< x< 16\\x\ne4\end{cases}}\)

\(\sqrt{x}=0\)  đúng 

Vì 

\(\sqrt{x}\ge0\forall x\)

\(\sqrt{x}=0\)  là đúng

=> Vì căn của 0 = 0

a) đk: \(2-m\ne0\Rightarrow m\ne2\)

b) Nếu \(m=4\)

\(\Rightarrow\left(2-4\right)x-4+1=0\)

\(\Leftrightarrow-2x-3=0\Leftrightarrow x=-\frac{3}{2}\)

Ta có: \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

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

\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

Đặt \(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(\Rightarrow A^2=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=\left(\sqrt{3-\sqrt{5}}\right)^2+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+\left(\sqrt{3+\sqrt{5}}\right)^2\)

\(=\left|3-\sqrt{5}\right|+2\sqrt{9-5}+\left|3+\sqrt{5}\right|\)

\(=3-\sqrt{5}+2\sqrt{4}+3+\sqrt{5}=6+4=10\)

Vì \(\hept{\begin{cases}\sqrt{3-\sqrt{5}}>0\\\sqrt{3+\sqrt{5}}>0\end{cases}}\Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}>0\)

=> A > 0 mà A² = 10 => A = √10

\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\)\(\frac{8}{1-\sqrt{5}}\)

\(=\frac{\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}\)\(+\frac{8}{1-\sqrt{5}}\)

\(=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}\)

\(=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\)

\(=-2\)

kb nhaaaaa

chúc bn hok tốt k mk nha