Các câu hỏi tương tự
Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)
a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
a, Cm công thức
\(\forall n\ge1\) ta có \(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng tính
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
Bài 1: CMR
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+........+\frac{1}{\left(n+1\right)\sqrt{n}}>2,n\varepsilonℕ^∗\)
Bài 2: Cho S= \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{3\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
CMR S<\(\frac{1}{2}\)
chứng minh \(\frac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}< \frac{1}{2}\)
Tính tổng sau:Afrac{1}{left[sqrt[3]{2}right]}+frac{1}{left[sqrt[3]{3}right]}+frac{1}{left[sqrt[3]{4}right]}+frac{1}{left[sqrt[3]{5}right]}+frac{1}{left[sqrt[3]{6}right]}+frac{1}{left[sqrt[3]{7}right]}+frac{1}{left[sqrt[3]{9}right]}+...+frac{1}{left[sqrt[3]{2012^3-1}right]}(trong tổng trên không có các số dạng frac{1}{left[sqrt[3]{n}right]} với n là lập phương 1 số nguyên,ví dụ:1 và 8)
Đọc tiếp
Tính tổng sau:
\(A=\frac{1}{\left[\sqrt[3]{2}\right]}+\frac{1}{\left[\sqrt[3]{3}\right]}+\frac{1}{\left[\sqrt[3]{4}\right]}+\frac{1}{\left[\sqrt[3]{5}\right]}+\frac{1}{\left[\sqrt[3]{6}\right]}+\frac{1}{\left[\sqrt[3]{7}\right]}+\frac{1}{\left[\sqrt[3]{9}\right]}+...+\frac{1}{\left[\sqrt[3]{2012^3-1}\right]}\)
(trong tổng trên không có các số dạng \(\frac{1}{\left[\sqrt[3]{n}\right]}\) với n là lập phương 1 số nguyên,ví dụ:1 và 8)
Bleft(frac{xsqrt{x}+x+sqrt{x}}{xsqrt{x}-1}-frac{sqrt{x}+3}{1-sqrt{x}}right).frac{x-1}{2x+sqrt{x}-1} ĐKXĐ: ...frac{left(xsqrt{x}+x+sqrt{x}right)left(1-sqrt{x}right)-left(sqrt{x}+3right)left(xsqrt{x}-1right)}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2x+2sqrt{x}-sqrt{x}-1}frac{xsqrt{x}+x+sqrt{x}-x^2-xsqrt{x}-x-x^2+sqrt{x}-3xsqrt{x}+3}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2sqrt{x}left(sqrt{x}+1right)-left(sqrt{x}+1right)}frac{-3xsqrt{x}+2sqrt{x}-2x^2+3}{left(xsqrt{x}-1ri...
Đọc tiếp
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
Rút gọnG frac{3-2sqrt{3}}{sqrt{3}}+frac{6}{3+sqrt{3}}H left(frac{1}{3-sqrt{5}}-frac{1}{3+sqrt{5}}right):frac{5-sqrt{5}}{sqrt{5}-1}i sqrt{frac{4}{left(2-sqrt{5}right)^2}}-sqrt{frac{4}{left(2+sqrt{5}right)^2}}K left(frac{2}{sqrt{3}-1}+frac{3}{sqrt{3}-2}+frac{15}{3-sqrt{3}}right).frac{1}{sqrt{3}+5}N left(1-frac{sqrt{3}-1}{2}right):left(frac{sqrt{3}-1}{2}+2right)
Đọc tiếp
Rút gọn
G = \(\frac{3-2\sqrt{3}}{\sqrt{3}}+\frac{6}{3+\sqrt{3}}\)
H= \(\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}\)
i = \(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
K = \(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
N= \(\left(1-\frac{\sqrt{3}-1}{2}\right):\left(\frac{\sqrt{3}-1}{2}+2\right)\)
16 tháng 9 2018 lúc 22:49
Với số tự nhiên n , \(n\ge3\)
Đặt \(S_n=\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
Chứng minh rằng \(S_n< \frac{1}{2}\)