\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\...">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

30 tháng 7 2018

ĐK:  \(a,b\ge0\);  \(a\ne b\)

\(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)

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

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

\(=a-b=VP\)

18 tháng 3 2019

ĐK: \(a,b\ge0,a\ne b\)

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

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

\(A=\left(\sqrt{a}+\sqrt{b}\right).\frac{1}{\sqrt{a}+\sqrt{b}}=1=VP\)

Vậy đẳng thức được cm.

5 tháng 4 2020

Ta có : \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)

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

\(=\frac{\left(\sqrt{a}^3+\sqrt{b}^3\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b}-\sqrt{ab}\)

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

\(=a-\sqrt{ab}+b-\sqrt{ab}\)

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

1,

\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)

24 tháng 9 2020

1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)

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

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

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

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

15 tháng 5 2017

Đặt cho dễ nhìn.

Đặt: \(\sqrt{a}=x\Rightarrow a=x^2;a\sqrt{a}=x^3\)

\(\sqrt{b}=y\Rightarrow b=y^2;b\sqrt{b}=y^3\)

\(\Leftrightarrow\frac{x^3+y^3}{x+y}-xy=\left(x-y\right)^2\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}-xy=x^2-2xy+y^2\)

\(\Leftrightarrow x^2-xy+y^2-xy=x^2-2xy+y^2\)

\(\Leftrightarrow x^2-2xy+y^2=x^2-2xy+y^2\)

\(\Rightarrowđpcm\)

24 tháng 9 2018

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

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

14 tháng 7 2019

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{\sqrt{a}-\sqrt{b}}\)

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

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

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

\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\right)-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{2\sqrt{a}.2\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{a}\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

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

\(=\frac{2\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{a-b}\)

Đề sai???Phân số thứ 3 nghi là a-b chứ ko phải căn a - căn b????????