Question

Cho A= \(\dfrac{\sqrt{x}}{\sqrt{x}+3}\) với ĐK x \(\ge\)0

B= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\) với ĐK x\(\ge\)0

Tính M biết M= A+B

Answer 1

\(M=A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}+3}=\dfrac{\sqrt{x}+2\sqrt{x}}{\sqrt{x}+3}=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\left(x\ge0\right)\)

Answer 2

`M=A+B`

`=sqrtx/(sqrtx+3)+(2sqrtx)/(sqrtx+3)`

`=(sqrtx+2sqrtx)/(sqrtx+3)`

`=(3sqrtx)/(sqrtx+3)`

Answer 3

Với \(x\ge0\), ta có:

\(M=A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}+3}\) \(=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\) \(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) \(=\dfrac{3x-9\sqrt{x}}{x-9}\)

#Cho mình sửa lại chút nhé! Nãy lag tí :)))