Bài 1: Căn bậc hai

\(\dfrac{\sqrt{x^6y^2}}{xy}=\dfrac{x^3y}{xy}=x^2\)

Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

Ta có: \(T=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)

=10-1

=9

Câu 2: 

Ta có: \(9-7x^2=30\)

\(\Leftrightarrow7x^2=9-30=-21\)(Vô lý)

a) \(x^2-9=\left(x-3\right)\left(x+3\right)\)

b) \(x^2+1-\dfrac{41}{25}=x^2-\dfrac{16}{25}=\left(x-\dfrac{4}{5}\right)\left(x+\dfrac{4}{5}\right)\)

`-5=-\sqrt(5^2)=-\sqrt25`

`-\sqrt29=-\sqrt29`

Vì `25<29`

`=> \sqrt25<\sqrt29`

`=> -\sqrt25 > -\sqrt29`

`=> -5 > -\sqrt29`

Vì \(25< 29\Rightarrow\sqrt{25}< \sqrt{29}\Rightarrow5< \sqrt{29}\Rightarrow-5>-\sqrt{29}\)

\(25< 29\\ \Leftrightarrow\sqrt{25}< \sqrt{29}\\ \Leftrightarrow-\sqrt{25}>-\sqrt{29}\\ \Leftrightarrow-5>-\sqrt{29}\)

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

b) \(4x-25=\left(2\sqrt{x}\right)^2-5^2=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4\left(x\ge0\right)\Rightarrow x+\sqrt{x}-4\sqrt{x}-4\)

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

b) \(4x-25=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

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

\(\sqrt{8-4\sqrt{3}}=\sqrt{6-2.\sqrt{6}.\sqrt{2}+2}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)

\(\sqrt{9-6\sqrt{2}}=\sqrt{6-2\sqrt{6}.\sqrt{3}+3}=\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}=\left|\sqrt{6}-\sqrt{3}\right|=\sqrt{6}-\sqrt{3}\)

\(\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}\)

\(\sqrt{9-6\sqrt{2}}=\sqrt{6}-\sqrt{3}\)

a) Ta có: \(\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{3}{\sqrt{7}-2}\right):\dfrac{2}{\sqrt{7}+\sqrt{3}}\)

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

\(=\dfrac{-\left(\sqrt{7}-\sqrt{3}\right)}{1}\cdot\dfrac{2}{\sqrt{7}-\sqrt{3}}\)

=-2

b) Ta có: \(\left(\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-1\right):\left(\sqrt{x}-x\right)+\dfrac{1}{x}\)

\(=\left(-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-1\right)\cdot\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{x}\)

\(=\left(-\sqrt{x}-1\right)\cdot\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{x}\)

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

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