Question

Rút gọn biêu thức:

c)\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

d)\(\sqrt{4\times\left(a-3\right)\times\left(b-2\right)}.\sqrt{9\times\left(b-3\right)^3\left(a-3\right)}\)

Mình nhầm câu d đoạn vế sau nha (b-2)³ mới đúng ạ

Mysterious PersonAkai Haruma

Answer 1

c) ta có : \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

d) ta có : \(\sqrt{4.\left(a-3\right)\left(b-2\right)}.\sqrt{9\left(b-2\right)^3\left(a-3\right)}\)

\(=\sqrt{36\left(a-3\right)^2\left(b-2\right)^4}=\left[{}\begin{matrix}6\left(a-3\right)\left(b-2\right)^2\left(nếu:a\ge3\right)\\6\left(3-a\right)\left(b-2\right)^2\left(nếu:a< 3\right)\end{matrix}\right.\)