Question

Rút gọn:

\(a,A=\sqrt{9\left(a+b\right)}-2\sqrt{16\left(a+b\right)}-3\sqrt{a+b}+\frac{1}{5}\sqrt{25\left(a+b\right)}\)

\(b,B=\frac{2ab\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{2ab\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(c,C=\frac{2ab}{a+\sqrt{ab}}+\frac{2ab}{b+\sqrt{ab}}\)

\(d,D=\frac{\frac{2ab\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{2ab\sqrt{b}}{\sqrt{a}+\sqrt{b}}}{\frac{2ab}{a+\sqrt{ab}}+\frac{2ab}{b+\sqrt{ab}}}\)

Giúp mình với.Thanks

Answer 1

tự làm

Answer 2

cho anh tao đê

Answer 3

a,A= -7\(\sqrt{a+b}\)

Answer 4

b,B= 2ab

Answer 5

a) \(A=\sqrt{9\left(a+b\right)}-2\sqrt{16\left(a+b\right)}-3\sqrt{a+b}+\frac{1}{5}\sqrt{25\left(a+b\right)}\)

\(A=3\sqrt{a+b}-8\sqrt{a+b}-3\sqrt{a+b}+\sqrt{a+b}\)

\(A=-7\sqrt{a+b}\)

b) ĐKXĐ : \(\hept{\begin{cases}\sqrt{a}+\sqrt{b}\ne0\\a\ge0\\b\ge0\end{cases}}\Leftrightarrow a\ne-b\ne0\)

 \(B=\frac{2ab\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{2ab\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(B=\frac{2ab\sqrt{a}+2ab\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)

Answer 6

c,C=2\(\sqrt{ab}\)

Answer 7

D=\(\sqrt{ab}\)