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CM : \(\sqrt{\left(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\right)^2}=1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
= \(\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=\frac{n^2\left[\left(n+1\right)^2+1\right]+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\) = \(\frac{n^2\left(n^2+2n+2\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
=\(\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\) = \(\frac{\left(n^2+n+1\right)^2}{\left(n^2+n\right)^2}\) =>\(\sqrt{\left(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\right)}=\frac{n^2+n+1}{n^2+n}\)
\(=1+\frac{1}{n^2+n}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có :
A = \(\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+\left(1+\frac{1}{4}-\frac{1}{5}\right)+...+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\)
= 2012 - \(\frac{1}{2013}\) \(\approx\) 2012
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
\(\forall k\ge0\)ta có :
\(\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\sqrt{k+1}-\sqrt{k}\)
Bạn áp dụng công thức này vào dãy trên ta sẽ có các số hạng triệt tiêu đi nhau và ra kết quả
\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}=\sqrt{100}-1=10-1=9\)
A = \(\frac{1}{1+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + . . . . . . . . . + \(\frac{1}{\sqrt{99+\sqrt{100}}}\)
= \(\sqrt{2}\) - 1 + \(\sqrt{2}\) - \(\sqrt{3}\) + . . . . . . . + \(\sqrt{100}\) - \(\sqrt{99}\)
= - 1 + \(\sqrt{100}\) = \(\sqrt{100}\) - 1 = 10 - 1 = 9
với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
=(\(\frac{1}{2}\sqrt{\frac{2}{2^2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{10^2.2}\)) .8
=(\(\frac{1}{2}\frac{\sqrt{2}}{2}-\frac{3\sqrt{2}}{2}+\frac{4.10\sqrt{2}}{5}\)).8
=(\(\frac{\sqrt{2}}{2}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\)).8
=\(2\sqrt{2}-12\sqrt{2}+64\sqrt{2}\)
=\(54\sqrt{2}\)