Question

rút gọn biểu thức: P=\(\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)

giúp mik!

Answer 1

`P=(x\sqrt{x}-3)/(x-2\sqrt{x}-3)-(2(\sqrt{x}-3))/(\sqrt{x}+1)+(\sqrt{x}+3)/(3-\sqrt{x})\  (ĐK:x\ge0;x\ne 9)`

`=(x\sqrt{x}-3)/((\sqrt{x}-3)(\sqrt{x}+1))-(2\sqrt{x}-6)/(\sqrt{x}+1)-(\sqrt{x}+3)/(\sqrt{x}-3)`

`=(x\sqrt{x}-3-(2\sqrt{x}-6)(\sqrt{x}-3)-(\sqrt{x}+3)(\sqrt{x}+1))/((\sqrt{x}-3)(\sqrt{x}+1))`

`=(x\sqrt{x}-3-(2x-12\sqrt{x}+18)-(x+4\sqrt{x}+3))/((\sqrt{x}-3)(\sqrt{x}+1))`

`=(x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3)/((\sqrt{x}-3)(\sqrt{x}+1))`

`=(x\sqrt{x}-3x+8\sqrt{x}-24)/((\sqrt{x}-3)(\sqrt{x}+1))`

`=(x(\sqrt{x}-3)+8(\sqrt{x}-3))/((\sqrt{x}-3)(\sqrt{x}+1))`

`=((\sqrt{x}-3)(\sqrt{x}+8))/((\sqrt{x}-3)(\sqrt{x}+1))`

`=(\sqrt{x}+8)/(\sqrt{x}+1)`

 

Answer 2

ĐKXĐ: \(x\ge0,x\ne9\)

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

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

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

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

\(=\dfrac{x\sqrt{x}-3-2x+4\sqrt{x}-18-x+3\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-3x+6\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

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