a)\(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}}{\sqrt{y}}.\left(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}+\sqrt{x}}\right)=\sqrt{\frac{x}{y}}\)
b) bí
 

tìm nghiệm hả bạn?

a) x^2-6=0

<=> x^2=6

<=> x=+- căn 6

b) x^2-5=0

<=> x^2=5

<=> x=+- căn 5

Ta có : x² - 5 = 0

=> x² = 5

=> x = \(-\sqrt{5}\); \(\sqrt{5}\)

a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

b. Khi \(x=2016-2\sqrt{2015}\Rightarrow A=\frac{2}{\sqrt{2016-2\sqrt{2015}}}\)

\(=\frac{2}{\sqrt{\left(\sqrt{2015}-1\right)^2}}=\frac{2}{\sqrt{2015}-1}\)

\(x^2-2x-2-2\sqrt{2x+1}=0\)

\(\Leftrightarrow x^2-2x-8-\left(2\sqrt{2x+1}-6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)-\frac{4\left(2x+1\right)-36}{2\sqrt{2x+1}+6}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)-\frac{8\left(x-4\right)}{2\sqrt{2x+1}+6}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2-\frac{8}{2\sqrt{2x+1}+6}\right)=0\)

Thấy: \(x+2-\frac{8}{2\sqrt{2x+1}+6}>0\)

\(\Rightarrow x-4=0\Rightarrow x=4\)