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16 tháng 8 2021

ĐK : a,b > 0

\(=\left[\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right]\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{a-b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

11 tháng 8 2018

\(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\div\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)

\(=\left(\frac{\sqrt{a}.\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)

\(=\left(\frac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a.\sqrt{a}.\left(\sqrt{b}-\sqrt{a}\right)+b.\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right).\left(b-a\right)}{\sqrt{ab}.\left(b-a\right)}\right)\)

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

11 tháng 8 2018

giải tiếp

\(=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a\sqrt{ab}+b\sqrt{ab}}{\sqrt{ab}\left(b-a\right)}\right)\)

\(=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{\sqrt{ab}.\left(a+b\right)}{\sqrt{ab}.\left(b-a\right)}\right)=\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}\right).\left(\frac{b-a}{a+b}\right)\)

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

20 tháng 8 2017

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

\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)

\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)

\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)


Điều kiện : a, b\(\ge0\)

6 tháng 6 2016

\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab-b}}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)

\(=\left(\frac{\sqrt{b}}{\sqrt{a}\sqrt{a}-\sqrt{a}\sqrt{b}}-\frac{\sqrt{a}}{\sqrt{a}\sqrt{b}-\sqrt{b}\sqrt{b}}\right).\left(\sqrt{a}\sqrt{a}\sqrt{b}-\sqrt{b}\sqrt{b}\sqrt{a}\right)\)

\(=\left(\frac{\sqrt{b}}{\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\frac{\left(\sqrt{b}\right)^2}{\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\left(\sqrt{a}\right)^2}{\sqrt{a}\sqrt{b}.\left(\sqrt{a}\sqrt{b}\right)}\right).\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{\left(\sqrt{b}\right)^2-\left(\sqrt{a}\right)^2}{\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)}.\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)\)

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

\(=b-a\)

7 tháng 12 2016

mi tích tau tau tích mi xong tau trả lời nka

 việt nam nói là làm

5 tháng 6 2017

bạn cứ quy đồng từ từ là ra đó

5 tháng 6 2017

mình thật sự muốn lm cho bạn nhưng nhìu việc quá,,,chỗ nào ko làm đc bạn hỏi mình,,,,mình làm cho

19 tháng 7 2019

\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2\) \(ĐKXĐ:\hept{\begin{cases}a\ge0\\b\ge0\\a\ne b\end{cases}}\)

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

\(=\left(\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\right)\left(\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)

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

\(=1\)

19 tháng 7 2019

\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

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

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

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

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

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

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

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

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

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

15 tháng 6 2017

\(=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)

\(=\frac{a-b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)

\(=a-b\)

\(\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right)\cdot\left(a\sqrt{b}-b\sqrt{a}\right)\)

\(=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\left[\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right]\)

\(=\frac{a-b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\left[\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right]\)

\(=\frac{a-b}{1}=a-b\)