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

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

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

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

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

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

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

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

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

b, \(\sqrt{8.32\left(2-a\right)^2}\)Với a =< 2 

\(=\sqrt{256\left(2-a\right)^2}=\sqrt{16^2\left(2-a\right)^2}=16\left|2-a\right|\)

\(=16\left(2-a\right)=32-16a\)

Tui nhầm đề xíu, cái A kia phải là:   A=\(\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}\)

thảo nào rút gọn mãi nó chả mất căn :))

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

\(=\sqrt{5}-1-\frac{5\sqrt{5}-10}{5}=\frac{5\sqrt{5}-5-5\sqrt{5}+10}{5}=\frac{5}{5}=1\)

Với \(x\ge0;x\ne4;9\)

\(P=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)

\(=\left(\frac{3\sqrt{x}+6+\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\right):\left(\sqrt{x}+3\right)\)

\(=\left(\frac{x+5\sqrt{x}+6}{x-4}\right):\left(\sqrt{x}+3\right)=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(x-4\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-2}\)

b, \(2P-A< 0\Rightarrow\frac{2}{\sqrt{x}-2}-1< 0\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-2}< 0\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-2}>0\)

TH1 : \(\hept{\begin{cases}\sqrt{x}-4>0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>16\\x>4\end{cases}\Leftrightarrow x>16}\)

TH2 : \(\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 16\\x< 4\end{cases}}\Leftrightarrow x< 4}\)

Kết hợp với đk vậy \(0\le x< 4;x>16\)

Tham khảo !
a + b)

c )