\(\sqrt{16+6\sqrt{7}}=a+\sqrt{7}\)

\(\Leftrightarrow\sqrt{9+2.3.\sqrt{7}+7}=a+\sqrt{7}\)

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

\(\Leftrightarrow3+\sqrt{7}=a+\sqrt{7}\)

\(\Leftrightarrow a=3\)

chúng ta cần làm gì

kmjkjpjpi

mình quên ĐK là x,y,z dương

6, \(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)\)

7, \(7+2\sqrt{10}=7+2\sqrt{5.2}=5+2\sqrt{5.2}+2=\left(\sqrt{5}+\sqrt{2}\right)^2\)

8, \(5-2\sqrt{6}=5-2\sqrt{2.3}=3-2\sqrt{2.3}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

9, \(\sqrt{x^2-y^2}-x+y=\sqrt{\left(x-y\right)\left(x+y\right)}-\left(x-y\right)\)

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

10, \(3x-2\sqrt{x}=\sqrt{x}\left(3\sqrt{x}-2\right)\)

1, \(\sqrt{xy}-x=\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)\)

2, \(x+y-2\sqrt{xy}=\left(\sqrt{x}-\sqrt{y}\right)^2\)

3, \(x\sqrt{y}-y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

4, \(2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}=2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)\)

\(=\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)\)

5, \(\sqrt{35}-\sqrt{14}=\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)\)

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

\(=2-x\cdot\frac{x-1}{x-2}=\frac{2x-4}{x-2}-\frac{x^2-x}{x-2}=\frac{-x^2+3x-4}{x-2}\)

\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)

Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)

Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)

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

Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)