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ta có: \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)
\(=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>1-\left(\frac{1}{2}-\frac{1}{101}\right)=1-\frac{1}{2}+\frac{1}{101}\)
\(=\frac{1}{2}+\frac{1}{101}\)
mà \(\frac{1}{2}=\frac{50}{100}>\frac{1}{100}\Rightarrow\frac{1}{2}+\frac{1}{101}>\frac{1}{100}\)
=> đ p c m
Ta có :
\(C=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=\left(\frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+...+\frac{n}{n!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\right)\)
\(=\left(1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{\left(n-1\right)!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....+\frac{1}{n!}\right)\)
\(=1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{\left(n-1\right)!}-\frac{1}{2!}-\frac{1}{3!}-\frac{1}{4!}-....-\frac{1}{n!}\)
\(=1-\frac{1}{n!}=\frac{n!-1}{n!}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
a) \(\frac{125^5}{5^{15}}=\frac{\left(5^3\right)^5}{5^{15}}=\frac{5^{15}}{5^{15}}=1\)
Mk không rảnh cho lắm !! nên chỉ làm câu a thui mấy câu khác để suy nghĩ đã
T nha
b) \(\left(\frac{2}{3}^{21}\right):\left(\frac{4}{9}^{10}\right)=\left(\frac{2}{3}^{21}\right):\left(\frac{2}{3}^2\right)^{10}=\left(\frac{2}{3}^{21}\right):\left(\frac{2}{3}^{20}\right)=\frac{2}{3}\)
\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-7}{20}\)
\(x=\frac{1}{4}:\frac{-7}{20}\)
\(x=\frac{-5}{7}\)
\(2x\left(x-\frac{1}{7}\right)=0\)
\(=>x=0\) hoặc \(x-\frac{1}{7}=0\)
\(x=0+\frac{1}{7}\)
\(x=\frac{1}{7}\)
Vậy \(x\in\left\{0;\frac{1}{7}\right\}\)
Mình làm câu b thôi nha
\(\frac{3}{4}\) + \(\frac{1}{4}\) : x = \(\frac{2}{5}\)
\(1\) : x = \(\frac{2}{5}\)
x = \(1\) * \(\frac{2}{5}\)
x = \(\frac{5}{2}\)
Answer : \(\frac{5}{2}\)
Ta có:
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}=\frac{1}{1\cdot2:2}+\frac{1}{2\cdot3:2}+...+\frac{1}{2012\cdot2013:2}\)
\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{2012\cdot2013}=2\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2012\cdot2013}\right]\)
\(=2\left[\left[\frac{1}{1}-\frac{1}{2}\right]+\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{4}\right]+...+\left[\frac{1}{2012}-\frac{1}{2013}\right]\right]\)
\(=2\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right]=2\left[1-\frac{1}{2013}\right]\)
\(=2\cdot\frac{2012}{2013}=\frac{4024}{2013}\)
Thế vào bài toán, ta có:
\(\frac{2\cdot2012}{\frac{4024}{2013}}=\frac{4024}{\frac{4024}{2013}}=2013\)