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a: \(1.5\sqrt{5}=\sqrt{1.5^2\cdot5}=\sqrt{\dfrac{45}{4}}\)
b: \(-ab^2\cdot\sqrt{5a}=-\sqrt{a^2b^4\cdot5a}=-\sqrt{5a^3b^4}\)
d: \(\dfrac{1}{3}y\sqrt{\dfrac{27}{y^2}}=-\sqrt{\dfrac{1}{9}y^2\cdot\dfrac{27}{y^2}}=-\sqrt{3}\)
c: \(\dfrac{1}{y}\sqrt{19y}=-\sqrt{\dfrac{1}{y^2}\cdot19y}=-\sqrt{\dfrac{19}{y}}\)
Bài 1:
c) \(\dfrac{1}{y}\sqrt{19y}=\sqrt{19y\cdot\dfrac{1}{y^2}}=\sqrt{\dfrac{19}{y}}\)
d) \(\dfrac{1}{3y}\cdot\sqrt{\dfrac{27}{y^2}}\cdot y=\sqrt{\dfrac{1}{9}\cdot\dfrac{27}{y^2}}=\sqrt{\dfrac{3}{y^2}}\)
Bài 3:
a) Ta có: \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{\sqrt{3}+5}\)
\(=\left(\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{1}+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right)\cdot\dfrac{1}{\sqrt{3}+5}\)
\(=\left(\sqrt{3}+1-2-\sqrt{3}+\dfrac{5\left(3+\sqrt{3}\right)}{2}\right)\cdot\dfrac{1}{\sqrt{3}+5}\)
\(=\left(-1+\dfrac{5\left(3+\sqrt{3}\right)}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\dfrac{-2+15+5\sqrt{3}}{2\left(5+\sqrt{3}\right)}\)
\(=\dfrac{13+5\sqrt{3}}{10+2\sqrt{3}}\)
Sửa đề: Đưa thừa số vào trong dấu căn
a: \(3\sqrt{x^2}=\sqrt{3^2\cdot x^2}=\sqrt{9x^2}\)
b: \(-5\sqrt{y^4}=-\sqrt{5^2\cdot y^4}=-\sqrt{25y^4}\)
c: \(3\sqrt{5x}=\sqrt{3^2\cdot5x}=\sqrt{45x}\)
d: \(x\sqrt{7}=\sqrt{x^2\cdot7}=\sqrt{7x^2}\)
a)\(=-\sqrt{\left(\frac{a}{b}\right)^2\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a^2}{b^2}\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a}{b}}\)
a: \(xy^2\sqrt{x}=\sqrt{x^2y^4\cdot x}=\sqrt{x^3y^4}\)
b: \(\dfrac{2}{x}\sqrt{\dfrac{15xy}{4}}=-\sqrt{\dfrac{4}{x^2}\cdot\dfrac{15xy}{4}}=-\sqrt{\dfrac{15y}{x}}\)
5 y y = ( 5 y ) 2 . y = 25. y 2 . y = 25. y 3
Chọn đáp án D