Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A.\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)\left(n+1-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. ap dungtinh B =\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Với mọi \(n\inℕ^∗\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
đến đây bạn áp dụng đẳng thức trên để tính gtbt nhé: kết quả: 9/10
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)
\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)
\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)
\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)}}+\frac{1}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
Với n thuộc N* ta luôn có :
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1^2+\left(\frac{1}{n}\right)^2+\left(\frac{1}{n+1}\right)^2+2.1.\frac{1}{n}-2.1.\frac{1}{n+1}-2.\frac{1}{n}.\frac{1}{n+1}}\)
\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}=1+\frac{1}{n}-\frac{1}{n+1}\)
Nên \(C=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\left(1+1+.....+1\right)+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99+\left(1-\frac{1}{100}\right)=100-\frac{1}{100}=\frac{9999}{100}\)
Bài 2:
a) \(\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{2-1}{\sqrt{1}+\sqrt{2}}=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}=\sqrt{2}-\sqrt{1}\)
Tương tự ta có: \(\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\);
\(\frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3}\); ............. ; \(\frac{1}{\sqrt{2024}+\sqrt{2025}}=\sqrt{2025}-\sqrt{2024}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{2025}-\sqrt{2024}\)
\(=\sqrt{2025}-\sqrt{1}=45-1=44\)
Bài 4:
\(M=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2.3.2\sqrt{2}+8}}-\frac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2.3.2\sqrt{2}+8}}\)
\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-\sqrt{8}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+\sqrt{8}\right)^2}}\)
\(=\frac{\left|\sqrt{2}-1\right|}{\left|3-\sqrt{8}\right|}-\frac{\left|\sqrt{2}+1\right|}{\left|3+\sqrt{8}\right|}=\frac{\sqrt{2}-1}{3-\sqrt{8}}-\frac{\sqrt{2}+1}{3+\sqrt{8}}\)
\(=\frac{\left(\sqrt{2}-1\right)\left(3+\sqrt{8}\right)}{\left(3-\sqrt{8}\right)\left(3+\sqrt{8}\right)}-\frac{\left(\sqrt{2}+1\right)\left(3-\sqrt{8}\right)}{\left(3+\sqrt{8}\right)\left(3-\sqrt{8}\right)}\)
\(=\left(3\sqrt{2}+\sqrt{16}-3-\sqrt{8}\right)-\left(3\sqrt{2}-\sqrt{16}+3-\sqrt{8}\right)\)
\(=3\sqrt{2}+4-3-\sqrt{8}-3\sqrt{2}+4-3+\sqrt{8}\)
\(=8-6=2\)là số tự nhiên