Tìm x biết:a) \(\frac{1}{5.8}\)+ \(\frac{1}{8.11}\)+ \(\frac{1}{11.14}\)+...+ \(\frac{1}{\left(x-2\right).x}\)+ \(\frac{1}{x\left(x+2\right)}\)= \(\frac{101}{1540}\).b) 1+ \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+...+ \(\frac{1}{x\left(x+1\right):2}\)=...
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Tìm x biết:
a) \(\frac{1}{5.8}\)+ \(\frac{1}{8.11}\)+ \(\frac{1}{11.14}\)+...+ \(\frac{1}{\left(x-2\right).x}\)+ \(\frac{1}{x\left(x+2\right)}\)= \(\frac{101}{1540}\).
b) 1+ \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+...+ \(\frac{1}{x\left(x+1\right):2}\)= 1\(\frac{1991}{1993}\).
a) Đặt \(A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{\left(x-2\right)x}+\frac{1}{x\left(x+2\right)}\)
=> \(3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{\left(x-2\right)x}+\frac{3}{x\left(x+2\right)}\)
=> \(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-2\right)}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+2}\)
=> 3A = \(\frac{1}{5}-\frac{1}{x+2}\)
=> A = \(\frac{1}{15}-\frac{1}{3x+6}\)
Mà : A = \(\frac{101}{1540}\)
=> \(\frac{1}{15}-\frac{1}{3x+6}=\frac{101}{1540}\)
=> \(\frac{1}{3x+6}=\frac{1}{15}-\frac{101}{1540}=\frac{1}{924}\)
=> 3x + 6 = 924
=> 3(x + 2) = 924
=> x + 2 = 308
=> x = 306
a) Ta có: \({{1} \over x(x+2)}= {{1} \over 3}({{1} \over x}-{{1} \over x+2})\) \(\Rightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over 8}+{{1} \over 8}-...+{{1} \over x}-{{1} \over x+2})={{101} \over 1540} \)\(\Leftrightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over x+2})={{101} \over 1540}\)\(\Leftrightarrow\)x+2 = 308 \(\Leftrightarrow\) x=306 Lúc sau lm hơi tắt mọi người thông cảm