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 

\(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}=4\sqrt{b}+2.2\sqrt{10b}-3.3\sqrt{10b}=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=4\sqrt{b}-5\sqrt{10b}\)

\(\sqrt{16b}+2\sqrt{40b}-30\sqrt{90b}\)

\(=\sqrt{16}\sqrt{b}+2\sqrt{40}\sqrt{b}-30\sqrt{90}\sqrt{b}\)

\(=\sqrt{b}\left(\sqrt{16}+2\sqrt{40}-30\sqrt{90}\right)\)

\(=\sqrt{b}\left(4+4\sqrt{10}-90\sqrt{10}\right)\)

\(=\sqrt{b}\left(4-86\sqrt{10}\right)\)

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

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

\(=\sqrt{b}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)

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

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

9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)

\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)

\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)

10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)

\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)

\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)

\(=-5\sqrt{3x}++27\)

11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)

\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)

\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)

\(=14\sqrt{2x}+28\)

12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)

\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)

\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)

\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)

\(=13\sqrt{5a}+\sqrt{a}\)

Lộn môn rồi nha bạn!

Với x ≥ 0; x ≠ 9 ta có:

\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x-3}\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

\(=\dfrac{3}{\sqrt{x}+3}\)

Vậy \(A=\dfrac{3}{\sqrt{x}+3}\).

\(A=\sqrt{4^2b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}=4\sqrt{b}+4\sqrt{10}\cdot\sqrt{b}-9\sqrt{10}\cdot\sqrt{b}\)

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

Rut gon A = √16b+2√40b−3√90bva`b≥0

 A=√42b+2√22·10b−3√32·10b=4√b+4√10·√b−9√10·√b

=4√b−5√10√b=(4−5√10)√b