cho \(a\ge0\). CMR:
\(\frac{a^2-\sqrt{a}}{a^2+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a^2-\sqrt{a}-1}+a+1=\left(\sqrt{a}-1\right)^2\)
K
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DK
2 tháng 8 2016
Đề của bạn bị sai, mình sửa lại đề ở dưới nhé!
\(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(a+1\right)\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(a+1\right)^2-\left(\sqrt{a}\right)^2}\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{\left(a-1\right)^2-a}\)
\(=\frac{a^3-a^2\sqrt{x}+a^2-a\sqrt{a}+a-\sqrt{a}-a^3-a^2\sqrt{a}-a^2-a\sqrt{a}-a-\sqrt{a}+\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{a^2+2a+1-a}\)
\(=\frac{-2a^2\sqrt{a}-2a\sqrt{a}-2\sqrt{a}+\left(a+1\right)\left(a^2+2a+1-a\right)}{a^2+a+1}\)
\(=\frac{-2\sqrt{a}\left(a^2+a+1\right)+\left(a+1\right)\left(a^2+a+1\right)}{a^2+a+1}\)
\(=\frac{\left(a^2+a+1\right)\left[-2\sqrt{x}+\left(x+1\right)\right]}{a^2+a+1}\)
\(=x-1-2\sqrt{x}\)
\(=\left(\sqrt{x}-1\right)^2\)
(Chúc you học giỏi nhoa!)
DK
2 tháng 8 2016
Ui ui mình ghi lộn, xin lỗi nhoa, vì x dễ làm hơn nên mình ghi lộn a thành x, mong bạn thông cảm hihi!
31 tháng 1 2022
\(VT=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+a+1\)
\(=a-\sqrt{a}-a-\sqrt{a}+a+1\)
\(=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2=VP\)
6 tháng 10 2020
a) Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\cdot\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\frac{1}{1+\sqrt{a}}\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{1+\sqrt{a}}\)
\(=1+\sqrt{a}\) Bằng 1 kiểu gì đây._.?
6 tháng 10 2020
a) Xin lỗi sửa lại phần a:
Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=1\)
b) Ta có: \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{6}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)=3-2=1\)
31 tháng 7 2020
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
LT
8 tháng 8 2019
A= (\(\frac{1}{\sqrt{a}-1}\) - \(\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\)) : \(\frac{\sqrt{a}-2}{a+1}\)
<=> (\(\frac{1}{\sqrt{a-1}}\) - \(\frac{2\sqrt{a}}{\left(a\sqrt{a-1}\right)-\sqrt{a}\left(\sqrt{a}-1\right)}\)). \(\frac{a+1}{\sqrt{a}-2}\)
<=> (\(\frac{1}{\sqrt{a}-1}\) - \(\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1-\sqrt{a}\right)}\)). \(\frac{a+1}{\sqrt{a}-2}\)
<=> (\(\frac{a+1}{\left(\sqrt{a}-1\right)\left(a+1\right)}\).\(\frac{a+1}{\sqrt{a}+2}\)
<=> \(\frac{a+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
\(VT=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+a+1\)
\(=a-\sqrt{a}-a-\sqrt{a}+a+1\)
\(=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2=VP\)