\(x^2-3x=2\sqrt{x-1}-4\)

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

\(\Leftrightarrow x^4-6x^3+17x^2-24x+16=4x-4\)

\(\Leftrightarrow x^4-6x^3+17x^2-24x+16-4x+4=0\)

\(\Leftrightarrow x^4-6x^3+17x^2-2x+20=0\)

\(\Leftrightarrow\left(x^3-4x^2+9x-10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2-2x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vì: \(x^2-2x+5\ne0\)

\(\Rightarrow x=2\)

\(x^2-4x+4+x-1-2\sqrt{x-1}+1=0\)

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

tự giải nốt

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\sqrt{20.9}+9}}}\)

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

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

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

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

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

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

\(\frac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}\)

\(=\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{5+2.2\sqrt{5}+4}\)

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

\(=\sqrt{5}-\sqrt{5}-2\)

\(=-2\)

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

=> \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)(1)

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

Từ (1) và (2) => \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}< \frac{1}{2\sqrt{n}}\)

\(\sqrt{14-8\sqrt{3}}\)\(=\sqrt{6-2.4.\sqrt{3}+8}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{3.16}+\left(\sqrt{8}\right)^2}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{48}+\left(\sqrt{8}\right)^2}\)

\(=\sqrt{\left(\sqrt{6}-\sqrt{8}\right)^2}\)

\(=\sqrt{6}-\sqrt{8}\)

1) \(x-y\)

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

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

2)\(1+x\sqrt{x}\)

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

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