Question

\(A=\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\) mấy bn giúp mik làm bài này với ,mik cảm ơn mấy bn nhiều!!!

Answer 1

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

(ĐKXĐ: x\(\ge\) 0 ; x \(\ne\) 1 )

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

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

\(=\sqrt{x}+1\)

Answer 2

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

Answer 3

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

\(A=\left(\sqrt{x}+1\right).\left(1-\sqrt{x}\right)\div\left(1-\sqrt{x}\right)\)\(A=\dfrac{1-x}{1-\sqrt{x}}\)