Question

cho M=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right)\cdot\dfrac{x-9}{\sqrt{9x}}\)

a, rút gọn M

b.tìm x để m bằng 6

Answer 1

`a)M=(sqrtx/(sqrtx-3)+sqrtx/(sqrtx+3))*(x-9)/sqrt{9x}(x>0,x ne 9)`

`M=((sqrtx(sqrtx+3)+sqrtx(sqrtx-3))/(x-9))*(x-9)/(3sqrtx)`

`M=((x+3sqrtx+x-3sqrtx)/(x-9))*(x-9)/(3sqrtx)`

`M=(2x)/(3sqrtx)=(2sqrtx)/3`

`b)M=6`

`=>2sqrtx=18`

`=>sqrtx=9=>x=81(tmđk)`

Answer 2

a) \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right).\dfrac{x-9}{\sqrt{9x}}\left(x>0,x\ne9\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{3\sqrt{x}}\)

\(=\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{3\sqrt{x}}=\dfrac{2\sqrt{x}}{3}\)

b) \(M=6\Rightarrow\dfrac{2\sqrt{x}}{3}=6\Rightarrow2\sqrt{x}=18\Rightarrow\sqrt{x}=9\Rightarrow x=81\)

Answer 3

a) Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right)\cdot\dfrac{x-9}{\sqrt{9x}}\)

\(=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{3\sqrt{x}}\)

\(=\dfrac{2x}{3\sqrt{x}}=\dfrac{2\sqrt{x}}{3}\)

b) Để M=6 thì \(\dfrac{2}{3}\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}=9\)

hay x=81(thỏa ĐK)