Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
Ta có công thức tổng quát
\(\dfrac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)Vậy \(P=\dfrac{1}{\sqrt{2}.1+\sqrt{1}.2}+\dfrac{1}{\sqrt{3}.2+\sqrt{2}.3}+...+\dfrac{1}{\sqrt{100}.99+\sqrt{99}.100}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Ta có công thức tổng quát
\(\dfrac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)Vậy \(P=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+...+\dfrac{1}{99\sqrt{100}+100\sqrt{99}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Em thử thôi chứ ko chắc đâu:((
Xét dạng tổng quát \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
Suy ra \(A=\frac{1}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{100-\sqrt{100}}{100}\)
b: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=-4\sqrt{5\sqrt{3}}\)
Ta có \(\dfrac{1}{\left(a+1\right)\sqrt{a}+a\sqrt{a+1}}\)
\(=\dfrac{1}{\sqrt{a\left(a+1\right)}\left(\sqrt{a+1}+\sqrt{a}\right)}\)
\(=\dfrac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a\left(a+1\right)}\left(\sqrt{a+1}+\sqrt{a}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}\)
\(=\dfrac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a\left(a+1\right)}}\)
(vì \(\left(\sqrt{a+1}+\sqrt{a}\right)\left(\sqrt{a+1}-\sqrt{a}\right)=\left(a+1\right)-a=1\))
\(=\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{a+1}}\)
Do đó \(B=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)