Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mình thật sự muốn lm cho bạn nhưng nhìu việc quá,,,chỗ nào ko làm đc bạn hỏi mình,,,,mình làm cho
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}\)
Tử số: \(\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+a\sqrt{b}+b\sqrt{a}+\sqrt{a}-\sqrt{b}-a\sqrt{b}+b\sqrt{a}}{1-ab}\)
\(=\frac{2\sqrt{a}+2\sqrt{a}b}{1-ab}=\frac{2\sqrt{a}\left(1+b\right)}{1-ab}\)
Mẫu số: \(1+\frac{a+b+2ab}{1-ab}=\frac{1-ab+a+b+2ab}{1-ab}=\frac{1+ab+a+b}{1-ab}=\frac{\left(a+b\right)\left(b+1\right)}{1-ab}\)
Do đó:
\(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right)}{\left(1+\frac{a+b+2ab}{1-ab}\right)}=\frac{\frac{2\sqrt{a}\left(1+b\right)}{1-ab}}{\frac{\left(a+1\right)\left(b+1\right)}{1-ab}}=\frac{2\sqrt{a}}{a+1}\)
ĐK: ab khác 1; a,b \(\ge\)0
\(B=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}:\frac{1-ab+a+b+2ab}{1-ab}\)
\(=\frac{2\sqrt{a}+2\sqrt{b}\sqrt{ab}}{1-ab}:\frac{1+ab+a+b}{1-ab}\)
\(=\frac{2\sqrt{a}\left(1+b\right)}{1-ab}:\frac{\left(1+b\right)\left(1+a\right)}{1-ab}\)
\(=\frac{2\sqrt{a}}{1+a}\)
https://vndoc.com/de-thi-hoc-sinh-gioi-mon-toan-lop-9-nam-hoc-2015-2016-truong-thcs-thanh-van-ha-noi/download
Trước hết ta rút gọn D :
\(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)(ĐKXĐ : \(a\ne0,b\ne0,ab\ne1\))
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}:\frac{1+a+b+ab}{1-ab}\)
\(=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}.\frac{1-ab}{\left(a+1\right)\left(b+1\right)}=\frac{2\sqrt{a}}{a+1}\)
a) Với \(a=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow D=\frac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{4-2\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}\)
b) Ta có : \(\left(\sqrt{a}-1\right)^2\ge0\Leftrightarrow a+1\ge2\sqrt{a}\Leftrightarrow\frac{2\sqrt{a}}{a+1}\le1\)
Suy ra Max D = 1 <=> a = 1
tự làm
cho anh tao đê