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\(P=\left(1+\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\frac{a+\sqrt{a}}{a-1}+\frac{\sqrt{a}}{a-\sqrt{a}}\right)\left(a>0;a\ne1\right)\)

\(P=\frac{a-\sqrt{a}+\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\)

\(P=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{\sqrt{a}-1}\right)\)

\(P=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

\(P=\frac{\sqrt{a}+1}{\sqrt{a}}:\frac{\sqrt{a}+1}{\sqrt{a}-1}\)

\(P=\frac{\sqrt{a}+1}{\sqrt{a}}.\frac{\sqrt{a}-1}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

Vậy .............

_Minh ngụy_

19 tháng 8 2019

\(P=\left(1+\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\frac{a+\sqrt{a}}{a-1}+\frac{\sqrt{a}}{a-\sqrt{a}}\right)\text{ (ĐKXĐ: }x\ne1;x\ne0\text{ )}\)

\(P=\left(\frac{a-\sqrt{a}}{a-\sqrt{a}}+\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\frac{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}{\left(a-1\right)\left(a-\sqrt{a}\right)}+\frac{\sqrt{a}\left(a-1\right)}{\left(a-\sqrt{a}\right)\left(a-1\right)}\right)\)

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

\(P=\frac{a-1}{a-\sqrt{a}}:\frac{a\left(a-1\right)+\sqrt{a}\left(a-1\right)}{\left(a-1\right)\left(a-\sqrt{a}\right)}\)

\(P=\frac{a-1}{a-\sqrt{a}}:\frac{\left(a-1\right)\left(a+\sqrt{a}\right)}{\left(a-1\right)\left(a-\sqrt{a}\right)}\)

\(P=\frac{a-1}{a-\sqrt{a}}:\frac{a+\sqrt{a}}{a-\sqrt{a}}\)

\(P=\frac{a-1}{a-\sqrt{a}}\times\frac{a-\sqrt{a}}{a+\sqrt{a}}\)

\(P=\frac{a-1}{a+\sqrt{a}}\)

30 tháng 7 2019

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

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)

\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)

\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)

\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)

\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)

NV
23 tháng 9 2019

ĐKXĐ:...

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

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

\(W=\left(\frac{\sqrt{a}-1}{a+\sqrt{a}+1}-\frac{a-3\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{1}{\sqrt{a}-1}\right).\left(\frac{1-\sqrt{a}}{a+1}\right)\)

\(=\left(\frac{\left(\sqrt{a}-1\right)^2-a+3\sqrt{a}-1-\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{1-\sqrt{a}}{a+1}\right)\)

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

7 tháng 12 2016

mi tích tau tau tích mi xong tau trả lời nka

 việt nam nói là làm

Cho e xin cảm ơn trc ak

5 tháng 11 2019

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

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

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

=\(1\)

9 tháng 7 2019

\(A=\)\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)

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

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

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

\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1-a\right)^2}=\frac{\sqrt{q}}{a+1}\)