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
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
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<\(\frac{1}{2}\)
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)
B1 : Rút gọn :6xy.sqrt{frac{9x^2}{16y^2}} left(x 0;yne0right)sqrt{frac{4+20a+25a^2}{b^4}}left(b 0;agefrac{-2}{5}right)left(m-nright).sqrt{frac{m-n}{left(m-nright)^2}}left(0 m nright)B2 : Tính :1.left(2sqrt{3}-sqrt{12}right):5sqrt{3}2.sqrt{frac{317^2-302^2}{1013^2-1012^2}}3.sqrt{27left(1-sqrt{3}right)^2}:3sqrt{75}4.left(5sqrt{frac{1}{5}}+frac{1}{2}sqrt{20}-frac{5}{4}sqrt{frac{4}{5}}+sqrt{5}right):2sqrt{5}
Đọc tiếp
B1 : Rút gọn :
\(6xy.\sqrt{\frac{9x^2}{16y^2}}\) \(\left(x< 0;y\ne0\right)\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}\)\(\left(b< 0;a\ge\frac{-2}{5}\right)\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}\)\(\left(0< m< n\right)\)
B2 : Tính :
\(1.\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}\)
\(2.\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}\)
\(3.\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(4.\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
1,cmr frac{2sqrt{mn}}{sqrt{n}+sqrt{n}+sqrt{m+n}}sqrt{m}+sqrt{n}-sqrt{m+n}1,rút gọn a, 3sqrt{27a}+2sqrt{frac{a}{3}}+asqrt{frac{4}{3a}}b,x^2sqrt{frac{12y}{x}}-xysqrt{frac{x}{3y}}c,frac{x}{left(sqrt{x}-sqrt{y}right)left(sqrt{x}-sqrt{z}right)}+frac{y}{left(sqrt{y}-sqrt{z}right)left(sqrt{y}-sqrt{x}right)}+frac{z}{left(sqrt{z}-sqrt{x}right)left(sqrt{z}-sqrt{y}right)}
Đọc tiếp
1,cmr
\(\frac{2\sqrt{mn}}{\sqrt{n}+\sqrt{n}+\sqrt{m+n}}\)=\(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)
1,rút gọn
a, 3\(\sqrt{27a}+2\sqrt{\frac{a}{3}}+a\sqrt{\frac{4}{3a}}\)
b,\(x^2\sqrt{\frac{12y}{x}}-xy\sqrt{\frac{x}{3y}}\)
c,\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)
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
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
rút gọn biểu thức B=\(\frac{1}{\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}}\)+\(\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}\)+....+\(\frac{1}{\sqrt[3]{n^2}+\sqrt[3]{n\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}\)
chứng minh \(\frac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}< \frac{1}{2}\)