Question

[cố quên một người]

a, Tìm x

(x+\(\dfrac{1}{2}\)) + (x+\(\dfrac{1}{6}\)) + (x+\(\dfrac{1}{12}\)) +..........+ (x+\(\dfrac{1}{9900}\)) =1

b, Tính tổng sau

A=\(3^2\)/1.4 + \(3^2\) /4.7 + \(3^2\)/7.10 +........+\(3^2\)/202.205

HELP ME......................! MAI MK PHẢI NỘP RÙI

Answer 1

[Adonis Baldric]

a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)

\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)

\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)

\(\Leftrightarrow50x+\dfrac{99}{100}=1\)

\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)

b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)

\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)

Answer 2

[Adonis Baldric]

a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)

Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)

\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)

\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)

\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)

\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)

b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)

\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)

\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)