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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: \(x\sqrt{7}=\sqrt{x^2\cdot7}=\sqrt{7x^2}\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
a: \(\sqrt{48a^4b^2}=\sqrt{16a^4b^2\cdot3}=4\sqrt{3}\cdot a^2\cdot\left|b\right|\)
\(=-4\sqrt{3}\cdot a^2b\)
b: \(\sqrt{-25x^3}=\sqrt{-25x^2\cdot x}=\left|25x^2\right|\cdot\sqrt{-x}\)
\(=-5x\sqrt{-x}\)
a: \(3\sqrt{200}=3\cdot10\sqrt{2}=30\sqrt{2}\)
b: \(-5\sqrt{50a^2b^2}=-5\cdot5\sqrt{2a^2b^2}\)
\(=-25\cdot\left|ab\right|\cdot\sqrt{5}\)
c: \(-\sqrt{75a^2b^3}\)
\(=-\sqrt{25a^2b^2\cdot3b}=-5\left|ab\right|\cdot\sqrt{3b}\)
a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)
b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)
\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)
( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))
c, Với \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
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: \(\dfrac{1}{y}\sqrt{19y}=-\sqrt{\dfrac{1}{y^2}\cdot19y}=-\sqrt{\dfrac{19}{y}}\)
a) \(2\sqrt{5a^2}=2\sqrt{5}\left|a\right|=-2a\sqrt{5}\)
b)\(2\sqrt{18a^2}=2.3\sqrt{2}.\left|a\right|=6a\sqrt{2}\)
c)\(\sqrt{-9b^3}=\sqrt{9.\left(-b\right)^3}=3\sqrt{-b}.\left|b\right|=-3b\sqrt{-b}\)
d)\(\sqrt{24a^4b^8}=\sqrt{6.\left(4a^2b^4\right)^2}=2a^2b^4\sqrt{6}\)
Lời giải:
a) $\sqrt{96.125}=\sqrt{96}.\sqrt{125}=\sqrt{4^2.6}.\sqrt{5^2.5}$
$=4\sqrt{6}.5\sqrt{5}=20\sqrt{30}$
b) $\sqrt{a^4.b^5}=\sqrt{(a^2b^2)^2.b}=a^2b^2\sqrt{b}$
c,d) Biểu thức không có căn.