Chứng minh đẳng thức :
\(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}}+\dfrac{1}{\sqrt{n}}\) ; với n thuộc N
Chứng minh đẳng thức sau \(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\) với n là một số tự nhiên tùy ý. Từ đó tính giá trị của biểu thức
\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}.\)
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 :
\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+.....+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}\)
\(=-1+10=9\)
Chứng minh đẳng thức :
\(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\) với n là số tự nhiên
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
Chứng minh các đẳng thức sau:
a) \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)
(Với \(x\ge0;x\ne1\))
b) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}+\dfrac{a-b}{\sqrt{a}-b}=2\sqrt{a}\)
(Với a>0; b>0; \(a\ne b\))
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Chứng minh: \(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}} = \dfrac{1}{(n+1).\sqrt{n} + n.\sqrt{n+1}}\)
Xét VT = \(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\left(\sqrt{n+1}\right)}\)
\(=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{n+1-n}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\sqrt{n+1}+\sqrt{n}.\sqrt{n}.\sqrt{n+1}}\)
\(=\dfrac{1}{\sqrt{n}.\left(n+1\right)+n.\sqrt{n+1}}\) = VP
=> Đpcm
chứng minh rằng với số tự nhiên n,n lớn hơn 4 ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}< 1\)
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
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)
Chứng minh: \(\sqrt{n}< \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n},\forall n\ge1\)
Giúp mình với mình đang cần gấp !
CM bất đẳng thức sau : \(2\sqrt{n+1}-2\sqrt{n}< \dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\sqrt{n-1}\)
\(\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}< \dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\\ \Leftrightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}\left(1\right)\\ \sqrt{n}-\sqrt{n-1}=\dfrac{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n-1}\right)}{\sqrt{n}+\sqrt{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}>\dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\\ \Leftrightarrow2\left(\sqrt{n}-\sqrt{n-1}\right)>\dfrac{1}{\sqrt{n}}\left(2\right)\)
\(\left(1\right)\left(2\right)\RightarrowĐfcm\)
Chứng minh: \(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n-1}+\sqrt{ }n}\)
Từ đó áp dụng tính \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}\)
\(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}\)
\(=\sqrt{6}-\sqrt{2}\)