Tất cả đều bằng nhau.

\(B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

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

\(=\frac{x+\sqrt{x}+12+3\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{4\sqrt{x}+12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(A=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{9}+\sqrt{8}\)

\(=\left|3-\sqrt{2}\right|-\sqrt{3^2}+\sqrt{2^2.2}\)

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

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

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

\(=\sqrt{2}\)

Mình cần gấp nha mn 😭😭 

1) Ta có: \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

 \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{81-17}=\sqrt{64}=8\)

\(\sqrt{36}+\sqrt{9}-\sqrt{49}\)

\(=6+3-7\)

\(=2\)

\(\sqrt{2}\cdot\left(\sqrt{50}-3\sqrt{2}\right)\)

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

\(=\sqrt{2}\cdot2\sqrt{2}\)

\(=4\)

a) \(T=\sqrt{36}+\sqrt{9}-\sqrt{49}\)

    \(=6+3-7\)

      \(=2\)

b)  \(B=\sqrt{2\left(\sqrt{50}-3\sqrt{2}\right)}\)

         \(=\sqrt{10\sqrt{2}-6\sqrt{2}}\)

           \(=\sqrt{\left(10-6\right)\sqrt{2}}\)

           \(=\sqrt{4\sqrt{2}}\)

            \(\approx2,39\)

\(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{9^2-\left(\sqrt{17}\right)^2}=\sqrt{81-17}\)

\(=\sqrt{64}=8.\)

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