Câu hỏi của Khánh Linh

Question

BT3: Tìm x, biết

20) $\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=\dfrac{1998}{2000}$

dao thi yen nhi · Accepted Answer

$\dfrac{1.2}{3.2}+\dfrac{1.2}{6.2}+.....+\dfrac{1.2}{2\left(x+1\right):2.2}$=$\dfrac{1998}{2000}$

$\dfrac{2}{6}+\dfrac{2}{12}+.....+\dfrac{2}{x\left(x+1\right)}$=$\dfrac{1998}{2000}$

$2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}$

$2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}$

$\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1998}{2000}:2$

$\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{999}{2000}$

$\dfrac{1}{x+1}=\dfrac{1}{2000}$

suy ra x+1=2000

suy ra x=2000-1=1999Câu trả lời: $\dfrac{1.2}{3.2}+\dfrac{1.2}{6.2}+.....+\dfrac{1.2}{2\left(x+1\right):2.2}$=$\dfrac{1998}{2000}$