ai lam ho voi

A=\(\sqrt{\left(4+\sqrt{8}\right)^2}\)\(-\sqrt{\left(4-\sqrt{8}\right)^2}\)=\(4+\sqrt{8}\)\(-\left(4-\sqrt{8}\right)\)=\(2\sqrt{8}\)

Giờ mình chỉ giải đc câu a thôi để hồi nao mình rảnh giải típ cho

a. Ta có \(A=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}+\frac{9}{\sqrt{x}-3}\)

\(=3+\frac{9}{\sqrt{x}-3}\)

\(A\in Z\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\Rightarrow\sqrt{x}-3\in\left\{-9;-3;-1;1;3;9\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\Rightarrow x\in\left\{0;4;16;36;144\right\}\)

Vậy \(x\in\left\{0;4;16;36;144\right\}\)thì \(A\in Z\)

b. Thay \(x=7-4\sqrt{3}\Rightarrow A=\frac{3\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}-3}\)

\(=\frac{3\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-3}=\frac{3\left(2-\sqrt{3}\right)}{2-\sqrt{3}-3}=\frac{15-9\sqrt{3}}{2}\)

\(=\frac{1}{\sqrt{1}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{957}+\sqrt{961}}\)

\(=\sqrt{5}-\sqrt{1}+\sqrt{9}-\sqrt{5}+...+\sqrt{961}-\sqrt{957}\)

\(=-\sqrt{1}+\sqrt{961}=\sqrt{961}-1=31-1=30\)

\(\hept{\begin{cases}\\\\\end{cases}\hept{\begin{cases}\\\end{cases}}\frac{ }{ }\sqrt[]{}\sqrt{ }\orbr{\begin{cases}\\\end{cases}}}\)

Ai lam ho voi minh dang can lam

ai làm hộ với mình đang cần gấp

a, ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(Q=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

b. \(Q>1\Rightarrow Q-1>0\Rightarrow\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Rightarrow\frac{x+2}{\sqrt{x}-1}>0\)

TH1 \(\hept{\begin{cases}x+2>0\\\sqrt{x}-1>0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x>1\end{cases}\Rightarrow}x>1}\)

TH2 \(\hept{\begin{cases}x+2< 0\\\sqrt{x}-1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\0\le x< 1\end{cases}\left(l\right)}}\)

Vậy \(x>1\)thì \(Q>1\)