Câu 18:
Ta có: \(3\sqrt{8a}+\dfrac{1}{4}\sqrt{\dfrac{32a}{25}}-\dfrac{a}{\sqrt{3}}\cdot\sqrt{\dfrac{3}{2a}}-\sqrt{2a}\)
\(=6\sqrt{2a}-\sqrt{2a}+\dfrac{1}{4}\cdot\dfrac{4\sqrt{2a}}{5}-\dfrac{a}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{2a}}\)
\(=5\sqrt{2a}+\dfrac{1}{5}\sqrt{2a}-\dfrac{1}{2}\sqrt{2a}\)
\(=\dfrac{47}{10}\sqrt{2a}\)
Chọn C
Câu 18
\(=3\sqrt{4}.\sqrt{2a}+\frac{1}{4}\sqrt{\frac{16}{25}}.\sqrt{2a}-\sqrt{\frac{a^2}{3}}.\sqrt{\frac{3}{2a}}-\sqrt{2a}\)
\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\sqrt{\frac{a}{2}}-\sqrt{2a}\)
\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\sqrt{\frac{1}{4}}.\sqrt{2a}-\sqrt{2a}\)
\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\frac{1}{2}\sqrt{2a}-\sqrt{2a}=\frac{47}{10}\sqrt{2a}\)
Đáp án C.
Câu 19:
\(=2\sqrt{a}-\sqrt{(3a)^2}.\sqrt{a}+a\sqrt{a}.\sqrt{16}+\sqrt{\frac{4}{a^4}.36a^5}\)
\(=2\sqrt{a}-3a\sqrt{a}+4a\sqrt{a}+\sqrt{144a}\)
\(=2\sqrt{a}+a\sqrt{a}+\sqrt{144}.\sqrt{a}=2\sqrt{a}+a\sqrt{a}+12\sqrt{a}=14\sqrt{a}+a\sqrt{a}\)
Đáp án A.
Câu 20:
\(=(\frac{1}{4}\sqrt{2^2.\frac{a}{2}}-\frac{3}{2}\sqrt{2a}+\frac{4}{5}.\sqrt{100}.\sqrt{2a}):\frac{1}{8}\)
\(=(\frac{1}{4}\sqrt{2a}-\frac{3}{2}\sqrt{2a}+8\sqrt{2a}).8=54\sqrt{2a}\)
Đáp án D.
