Câu hỏi của cố quên một người

Question

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 +........+$3^2$/2...

Adonis Baldric · Accepted Answer

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

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

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

$A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}$Câu trả lời: 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) $A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}$

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

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

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

Adonis Baldric · Answer

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

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

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

$A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}$Câu trả lời: 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) $A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}$

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

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

$A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}$

cố quên một người · Answer

dấu gạch chéo của phép tính b là ''phần'' các bn nhé.Câu trả lời: dấu gạch chéo của phép tính b là ''phần'' các bn nhé.

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