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\(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
VP = \(\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
= \(\sqrt{n+1}-\sqrt{n}\)
= VT
Vậy đẳng thức được chứng minh
Ta có : \(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{1}\)
\(=\sqrt{n+1}-\sqrt{n}\)
Vậy đẳng thức đã được chứng minh .
Áp dụng :
\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+.....+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}\)
\(=-1+10=9\)
\(\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{\left(n+1\right)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\) (đpcm)
\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)+n=2n+1=\left(n+1-n\right)\left(n+1+n\right)=\left(n+1\right)^2-n^2\)
Sửa đề:
\(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\\ < =>\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)=1\\ < =>n+1-n=1\\ < =>1=1\)(luôn luôn đúng)
=> đfcm
biến đổi vế phải ta có :
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\left(đpcm\right)\)
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó:
\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)
a)= \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
= \(-1+\sqrt{100}\)
= -1 +10
=9
b)Ta có\(\left(\sqrt{n+1}-\sqrt{n}\right)\cdot\left(\sqrt{n+1}+\sqrt{n}\right)\)=n+1-n=1 (1)
Lại có:\(\frac{1}{\sqrt{n+1}+1}\cdot\left(\sqrt{n+1}+1\right)=1\)(2)
Từ (1) và (2)=>\(\left(\sqrt{n+1}-1\right)=\frac{1}{\sqrt{n+1}+1}\)
Ta có: \(\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)=\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2=n+1-n=1\) \(\Leftrightarrow\) \(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\) với n là số tự nhiên