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13 tháng 9 2021

Với x > = 0 ; \(x\ne1\)

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

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

\(=\frac{2\left(2\sqrt{x}\right)}{\left(1-x\right)^2}+\frac{1}{1-\sqrt{x}}=\frac{4\sqrt{x}}{\left(x-1\right)^2}-\frac{\left(\sqrt{x}+1\right)\left(x-1\right)}{\left(x-1\right)^2}\)

\(=\frac{4\sqrt{x}-\left(\sqrt{x}+1\right)\left(x-1\right)}{\left(x-1\right)^2}\)

24 tháng 8 2017

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

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

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

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 6 2019

ĐK : x>0, x khác 1

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

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

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

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

29 tháng 8 2020

ĐKXĐ: \(x>0;x\ne1\)

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

27 tháng 1 2020

;))) tớ nhớ dạng RGBT căn bậc 3 lớp 9 nhì :)))???? 

\(\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}}{x+\sqrt{x+1}}\right).\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(=\frac{2x+1-\sqrt{x}\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left[\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)

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

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

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