$\frac{x}{2}$ +$\frac{x}{6}$ +$\frac{x}{12}$ +…+$\frac{x}{9702}$ =9898 có ai giúp mik với 22/09/2021 Bởi aikhanh $\frac{x}{2}$ +$\frac{x}{6}$ +$\frac{x}{12}$ +…+$\frac{x}{9702}$ =9898 có ai giúp mik với
$\begin{array}{l}\dfrac x2+\dfrac x6+\dfrac x{12}+\dots+\dfrac x{9702}=9898\\\Leftrightarrow x\cdot\left(\dfrac12+\dfrac16+\dfrac1{12}+\dots+\dfrac1{9702}\right)=9898\\\Leftrightarrow x\cdot\left(\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}+\dots+\dfrac1{98.99}\right)=9898\\\Leftrightarrow x\cdot\left(1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dots+\dfrac1{98}-\dfrac1{99}\right)=9898\\\Leftrightarrow x\cdot\left(1-\dfrac1{99}\right)=9898\\\Leftrightarrow x\cdot\dfrac{98}{99}=9898\\\Leftrightarrow x=9898\div\dfrac{98}{99}\\\Leftrightarrow x=\dfrac{9898.99}{98}\\\Leftrightarrow x=\dfrac{98.101.99}{98}\\\Leftrightarrow x=101.99\\\Leftrightarrow x=9999 \end{array}$ Bình luận
Đáp án: `x=9999` Giải thích các bước giải: $\rm\dfrac{x}{2}+\dfrac{x}{6}+\dfrac{x}{12}+…+\dfrac{x}{9702}=9898\\⇒x . \dfrac{1}{2}+x . \dfrac{1}{6}+x . \dfrac{1}{12}+…+ x . \dfrac{1}{9702}=9898\\⇒x.\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+…+\dfrac{1}{9702}\right)=9898\\⇒x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{98.99}\\⇒x.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{98}-\dfrac{1}{99}\right.\right)=9898\\⇒x.\left(\dfrac{1}{1}-\dfrac{1}{99}\right)=9898\\⇒x.\left(\dfrac{99}{99}-\dfrac{1}{99}\right)=9898\\⇒x.\dfrac{99-1}{99} =9898\\⇒x.\dfrac{98}{99}=9898\\⇒x=9898:\dfrac{98}{99}\\⇒x=9898.\dfrac{99}{98}\\⇒x=\dfrac{9898.99}{98}\\=>x=\dfrac{98.101.99}{98}\\⇒x=\dfrac{1.101.99}{1}\\⇒x=101.99\\⇒x=9999\\Vậy \ x \ = \ 9999$ Bình luận
$\begin{array}{l}\dfrac x2+\dfrac x6+\dfrac x{12}+\dots+\dfrac x{9702}=9898\\\Leftrightarrow x\cdot\left(\dfrac12+\dfrac16+\dfrac1{12}+\dots+\dfrac1{9702}\right)=9898\\\Leftrightarrow x\cdot\left(\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}+\dots+\dfrac1{98.99}\right)=9898\\\Leftrightarrow x\cdot\left(1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dots+\dfrac1{98}-\dfrac1{99}\right)=9898\\\Leftrightarrow x\cdot\left(1-\dfrac1{99}\right)=9898\\\Leftrightarrow x\cdot\dfrac{98}{99}=9898\\\Leftrightarrow x=9898\div\dfrac{98}{99}\\\Leftrightarrow x=\dfrac{9898.99}{98}\\\Leftrightarrow x=\dfrac{98.101.99}{98}\\\Leftrightarrow x=101.99\\\Leftrightarrow x=9999 \end{array}$
Đáp án:
`x=9999`
Giải thích các bước giải:
$\rm\dfrac{x}{2}+\dfrac{x}{6}+\dfrac{x}{12}+…+\dfrac{x}{9702}=9898\\⇒x . \dfrac{1}{2}+x . \dfrac{1}{6}+x . \dfrac{1}{12}+…+ x . \dfrac{1}{9702}=9898\\⇒x.\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+…+\dfrac{1}{9702}\right)=9898\\⇒x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{98.99}\\⇒x.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{98}-\dfrac{1}{99}\right.\right)=9898\\⇒x.\left(\dfrac{1}{1}-\dfrac{1}{99}\right)=9898\\⇒x.\left(\dfrac{99}{99}-\dfrac{1}{99}\right)=9898\\⇒x.\dfrac{99-1}{99} =9898\\⇒x.\dfrac{98}{99}=9898\\⇒x=9898:
\dfrac{98}{99}\\⇒x=9898.\dfrac{99}{98}\\⇒x=\dfrac{9898.99}{98}\\=>x=\dfrac{98.101.99}{98}\\⇒x=\dfrac{1.101.99}{1}\\⇒x=101.99\\⇒x=9999\\Vậy \ x \ = \ 9999$