2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)

\(=\left(7-6+1\right)\sqrt{2}\)

\(=2\sqrt{2}\)

3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)

\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)

\(=\left(3-4+7\right)\sqrt{a}\)

\(=6\sqrt{a}\)

4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)

\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

\(=4\sqrt{b}-5\sqrt{10b}\)

Gấp nha 

Mẫu số bằng \(\sqrt{50}+\sqrt{250}=5\sqrt{2}+5\sqrt{10}=5\sqrt{2}\left(1+\sqrt{5}\right).\)

Kí hiệu tử số là \(A\) thì ta có 

\(A^2=\left(\sqrt{8+\sqrt{40+8\sqrt{5}}}+\sqrt{8-\sqrt{40+8\sqrt{5}}}\right)^2\)

       \(=8+\sqrt{40+8\sqrt{5}}+2\sqrt{8+\sqrt{40+8\sqrt{5}}}\cdot\sqrt{8-\sqrt{40+8\sqrt{5}}}+8-\sqrt{40+8\sqrt{5}}\)

       \(=16+2\sqrt{\left(8+\sqrt{40+8\sqrt{5}}\right)\left(8-\sqrt{40+8\sqrt{5}}\right)}\)

      \(=16+2\sqrt{8^2-\left(40+8\sqrt{5}\right)}=16+2\sqrt{24-8\sqrt{5}}\)

      \(=16+2\sqrt{4-2\cdot2\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}=16+2\sqrt{\left(2-2\sqrt{5}\right)^2}\)

      \(=16+2\left|2-2\sqrt{5}\right|=16-4+4\sqrt{5}=12+4\sqrt{5}=4\left(3+\sqrt{5}\right).\)

Vậy  \(A=4\left(3+\sqrt{5}\right)=2\left(6+2\sqrt{5}\right)=2\left(\sqrt{5}+1\right)^2.\)

Thành thử  \(B=\frac{2\left(\sqrt{5}+1\right)^2}{5\sqrt{2}\left(1+\sqrt{5}\right)}=\frac{\sqrt{2}\left(\sqrt{5}+1\right)}{5}=\frac{\sqrt{10}+\sqrt{2}}{5}.\)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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