Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{72a^8\left(x^2-4x+4\right)}=\sqrt{72a^8\left(x-2\right)^2}=\sqrt{72}a^4|\left(x-2\right)|=\sqrt{72}a^4\left(2-x\right)\)
\(\sqrt{40x^6\left(a^2+6a+9\right)}=\sqrt{40x^6\left(x+3\right)^2}=\sqrt{40}|x^3\left(x+3\right)|=\sqrt{40}.\left(-x^3\right)\left(3-x\right)\)
\(=-\sqrt{40}x^3\left(3-x\right)\)
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
\(=5x\left(2y-1\right)^2=5.\left(-\sqrt{2}\right)\left(6-1\right)^2=-\sqrt{2}.5^3\)