Cho biểu thức : \(C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) CMR: \(C< \dfrac{3}{16}\)
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Ta có: \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\cdot\left(1-2y\right)=40\)
Xét ta có: \(1-2y\) lẻ với mọi y
Mà: Ư(40)=\(\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
\(\Rightarrow1-2y\in\left\{\pm1;\pm5\right\}\)
\(1-2y\) | -1 | 1 | 5 | -5 |
\(x\) | -40 | 40 | 8 | -8 |
\(y\) | 1 | 0 | -2 | 3 |
Vậy: ...
Bài 5:
a) Góc so le trong với góc BAn: \(\widehat{ABC}\)
Góc so le trong với góc nAC: \(\widehat{ACy}\)
b) Góc trong cùng phía với góc B là: \(\widehat{BAM}\)
Góc trong cùng phía với góc ACB là: \(\widehat{nAC}\)
\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)
\(B=\left(\dfrac{2^2}{2^2}-\dfrac{1}{2^2}\right)\cdot\left(\dfrac{3^2}{3^2}-\dfrac{1}{3^2}\right)....\left(\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\right)\)
\(B=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}....\cdot\dfrac{100^2-1}{100^2}\)
\(B=\dfrac{\left(2+1\right)\left(2-1\right)}{2^2}\cdot\dfrac{\left(3+1\right)\left(3-1\right)}{3^2}\cdot...\cdot\dfrac{\left(100+1\right)\left(100-1\right)}{100^2}\)
\(B=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}\)
\(B=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot101}{2^2\cdot3^2\cdot4^2\cdot5^2\cdot....\cdot100^2}\)
\(B=\dfrac{1\cdot101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)
\(B=\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)
Mà: \(\dfrac{1}{2}=\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)
Ta có: \(101< 3\cdot4\cdot5\cdot...\cdot100\)
\(\Rightarrow\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}< \dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)
\(\Rightarrow B< \dfrac{1}{2}\)
\(a,101^2=101.\left(100+1\right)=10100+101=10201.\\ b,75^2-50.75+25^2\\ =75.\left(75-50\right)+25^2\\ =75.25+25^2\\ =25.\left(75+25\right)\\ =25.100\\ =2500.\)
\(c,103.97\\ =\left(100+3\right).97\\ =9700+291\\ =9991\)
Đặt:
\(A=\dfrac{7}{11\cdot16}+\dfrac{7}{16\cdot21}+\dfrac{7}{21\cdot26}+...+\dfrac{7}{61\cdot66}\)
\(\dfrac{5}{7}A=\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{61\cdot66}\)
\(\dfrac{5}{7}A=\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\)
\(\dfrac{5}{7}A=\dfrac{1}{11}-\dfrac{1}{66}=\dfrac{6}{66}-\dfrac{1}{66}=\dfrac{5}{66}\)
\(A=\dfrac{5}{66}\cdot\dfrac{7}{5}=\dfrac{7}{66}\)
A = \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)
A = \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
A = - \(\dfrac{3}{5}\)
a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\)\(x\) = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\) \(x\) = - \(\dfrac{3}{20}\)
\(x\) = - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)
\(x\) = - \(\dfrac{3}{8}\)
b, \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)
\(x\) = \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))
\(x\) = - \(\dfrac{15}{16}\)