Question

giải giúp mình bài 9 câu C , Q với ạ :<

Answer 1

Bài 9:

\(Q=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(C=\dfrac{x-2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

Answer 2

c, đk x >= 0 ; x khác 1 

\(C=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{-\sqrt{x}+1+x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{x\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

q, đk x >= 0 ; x khác 4 ; 9 

\(Q=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-5\sqrt{x}+6}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{x-5\sqrt{x}+6}=\dfrac{x-\sqrt{x}-2}{x-5\sqrt{x}+6}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Answer 3

Với `x >= 0,x \ne 4,x \ne 9` có:

`Q=[2\sqrt{x}-9]/[x-5\sqrt{x}+6]-[\sqrt{x}+3]/[\sqrt{x}-2]-[2\sqrt{x}+1]/[3-\sqrt{x}]`

`Q=[2\sqrt{x}-9-(\sqrt{x}+3)(\sqrt{x}-3)+(2\sqrt{x}+1)(\sqrt{x}-2)]/[(\sqrt{x}-3)(\sqrt{x}-2)]`

`Q=[2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2]/[(\sqrt{x}-3)(\sqrt{x}-2)]`

`Q=[x-\sqrt{x}-2]/[(\sqrt{x}-3)(\sqrt{x}-2)]`

`Q=[(\sqrt{x}-2)(\sqrt{x}+1)]/[(\sqrt{x}-3)(\sqrt{x}-2)]`

`Q=[\sqrt{x}+1]/[\sqrt{x}-3]`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Với `x >= 0,x \ne 1` có:

`C=[x+2]/[x\sqrt{x}-1]+\sqrt{x}/[x+\sqrt{x}+1]-1/[\sqrt{x}-1]`

`C=[x+2+\sqrt{x}(\sqrt{x}-1)-x-\sqrt{x}-1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`

`C=[x+2+x-\sqrt{x}-x-\sqrt{x}-1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`

`C=[x-2\sqrt{x}+1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`

`C=[(\sqrt{x}-1)^2]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`

`C=[\sqrt{x}-1]/[x+\sqrt{x}+1]`