Question

a)\(\sqrt{\dfrac{a^2}{25+10b+b^2}}\) với a < 0, b >0

b)\(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}\)với a khác b

c)\(\dfrac{x+4\sqrt{x}+4}{2+\sqrt{x}}\)với x >= 0

 

Answer 1

(a) \(\sqrt{\dfrac{a^2}{25+10b+b^2}}=\sqrt{\dfrac{a^2}{\left(5+b\right)^2}}=\dfrac{\sqrt{a^2}}{\sqrt{\left(5+b\right)^2}}\)

\(=\dfrac{\left|a\right|}{\left|5+b\right|}=\dfrac{-a}{b+5}\) (do \(a< 0,b>0\Rightarrow b+5>0\))

 

(b) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\sqrt{\dfrac{\left(ab\right)^2}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}\)

\(=\left(a-b\right)\cdot\dfrac{\left|ab\right|}{\left|a-b\right|}\).

 

(c) \(\dfrac{x+4\sqrt{x}+4}{2+\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}=\sqrt{x}+2.\)