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

\(=\frac{\sqrt{3}+\sqrt{2}-1}{3+2\sqrt{3}\sqrt{2}+2-1}=\frac{\sqrt{3}+\sqrt{2}-1}{2\sqrt{3}\sqrt{2}+4}=\frac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(2\sqrt{3}\sqrt{2}-4\right)}{4.3.2-16}\)

\(=\frac{2.3.\sqrt{2}-4.\sqrt{3}+2.2.\sqrt{3}-4.\sqrt{2}-2.\sqrt{3}.\sqrt{2}+4}{8}=\)

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

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

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