Nhầm nha mn 

\(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)

\(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)

\(=\frac{5+4+4\sqrt{5}-8\sqrt{5}}{2\sqrt{5}-4}\)

\(=\frac{9-4\sqrt{5}}{2\sqrt{5}-4}\)

\(=\frac{\sqrt{5}^2-4\sqrt{5}+2^2}{2\sqrt{5}-4}\)

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

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

Với \(x\ge\)0

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

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

ta có :

\(A=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}-1\ge2-1=1\)

Vì A>1 nên \(A>\sqrt{A}\)

ta có :

\(P=\sqrt{\sqrt[3]{x^4}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}+\sqrt{\sqrt[3]{y^4}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}\)\(\)

\(=\sqrt[3]{x^2}.\sqrt{\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}+\sqrt[3]{y^2}.\sqrt{\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}=\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right).\sqrt{\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}\)

\(=\sqrt{\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)^3}\Leftrightarrow\sqrt[3]{P^2}=\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)\)

Vậy ta có đpcm

\(\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}{2+3-5+2\sqrt{6}}\)

\(=\frac{2\sqrt{6}\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}{2\sqrt{6}}=\sqrt{2}+\sqrt{3}-\sqrt{5}\)

Điều kiên \(0\le x\le1\)

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

ta có :\(\left(\sqrt{x}+\sqrt{1-x}\right)^2=1+2\sqrt{x}.\sqrt{1-x}\ge1\)

và \(\sqrt{x+1}\ge1\) với \(x\ge0\)Vậy

\(\sqrt{x}+\sqrt{x+1}+\sqrt{1-x}\ge2\)

dấu bằng xảy ra khi x=0 

vậy x=0 là nghiệm duy nhất của phương trình

\(\left|a+b-\frac{a}{\frac{a}{b}+1}\right|=\sqrt{\left(a+b-\frac{a}{\frac{a}{b}+1}\right)^2}=\sqrt{\left(a+b\right)^2-\frac{2\left(a+b\right)a}{\frac{a}{b}+1}+\frac{a^2}{\left(\frac{a}{b}+1\right)^2}}\)

\(=\sqrt{\left(a+b\right)^2-2ab+\frac{a^2}{\left(\frac{a}{b}+1\right)^2}}=\sqrt{a^2+b^2+\frac{a^2}{\left(\frac{a}{b}+1\right)^2}}\)

\(\hept{\begin{cases}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{2}x+\sqrt{2}y=\sqrt{2}+1\\x=1-\left(\sqrt{2}+1\right)y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}1-\left(\sqrt{2}+1\right)y+y=\frac{2+\sqrt{2}}{2}\\x=1-\left(\sqrt{2}+1\right)y\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=\frac{3+\sqrt{2}}{2}\end{cases}}\)

\(\frac{3}{4}=tanB=\frac{AC}{AB}\Rightarrow AC=\frac{3}{4}AB\)

\(BC^2=AB^2+AC^2\)(định lí Pythagore) 

\(=AB^2+\frac{9}{16}AB^2=\frac{25}{16}AB^2\)

\(\Rightarrow AB^2=10^2.\frac{16}{25}\Rightarrow AB=8\left(cm\right)\)

\(\Rightarrow AC=\frac{3}{4}AB=6\left(cm\right)\)