Rút gọn
\(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\)
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24 tháng 7 2019
ĐKXĐ: Bạn tự làm nha
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+1\)
\(=\frac{x^2-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x^2+x+1}{x+\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1\left(\sqrt{a}-1\right)-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1-2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
29 tháng 1 2022
\(C=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\sqrt{x}-1}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
O
4 tháng 10 2019
A=(\(\frac{x-2}{x+2\sqrt{x}}\)+\(\frac{1}{\sqrt{x}+2}\)).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=(\(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)+\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=\(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=\(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
A=\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy............................
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\sqrt{x}^2-1}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2}{\sqrt{x}^2-1}=\frac{2}{x-1}\)