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Chứng minh công thức:
\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^2\left(a^2+2a+1\right)+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^4+2a^3+a^2+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^4+2a^3+3a^2+2a+1}{a^2\left(a+1\right)^2}}\)
=\(\sqrt{\dfrac{\left(a^2\right)^2+2a^2a+2a^2+2a+a^2+1}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}}\)
\(=\dfrac{a^2+a+1}{a\left(a+1\right)}\)
\(=\dfrac{a\left(a+1\right)+1}{a\left(a+1\right)}\)
\(=1+\dfrac{1}{a\left(a+1\right)}\)
\(=1+\dfrac{1}{a}-\dfrac{1}{a+1}\)
Áp dụng công thức ta có:
\(C=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{1999^2}+\dfrac{1}{2000^2}}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=2000-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)
\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
=>A là số hữu tỉ (ĐPCM)
\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)
\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)
bạn chứng minh bài toán tổng quát : \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=1+\frac{1}{a}-\frac{1}{a+1}\)rồi áp dụng vào giải bài này nhé
b: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=-4\sqrt{5\sqrt{3}}\)
1: \(=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
2: \(=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}=\dfrac{8}{2}=4\)
4: \(=\dfrac{-3+5\sqrt{3}}{11}+\dfrac{3+5\sqrt{3}}{11}=\dfrac{10\sqrt{3}}{11}\)
Ta chứng minh công thức:
\(1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}=\left(1+\dfrac{1}{n}+\dfrac{1}{n+1}\right)^2\) bằng cách quy đồng biểu thức ở vế phải rồi áp dụng vào bài tập