Câu hỏi của Trần Châu

Question

(x+1) +(x+3) +(x+5)+... +(x+99) =1000

Trần Đình Thiên · Accepted Answer

(x+1)+(x+3)+...+(x+99)=1000

=>50x+1+3+...+99=1000

=>50x+$\dfrac{\left(99-1\right):2+1\cdot\left(1+99\right)}{2}$=1000

=>50x+2500=1000

=>50x=1000-2500=-1500

=>x=-1500:50=-30.Câu trả lời: (x+1)+(x+3)+...+(x+99)=1000

=>50x+1+3+...+99=1000

=>50x+$\dfrac{\left(99-1\right):2+1\cdot\left(1+99\right)}{2}$=1000

=>50x+2500=1000

=>50x=1000-2500=-1500

=>x=-1500:50=-30.

Nguyễn Nhân Dương · Answer

$\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=1000$

$\left(x+x+x+...+x\right)+\left(1+3+5+7+9+...+99\right)$

$\left(x.50\right)+2500$=1000

$x.50=1000-2500$

$x.50=-1500$

$x=-1500:50=-30$Câu trả lời: $\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=1000$

$\left(x+x+x+...+x\right)+\left(1+3+5+7+9+...+99\right)$

$\left(x.50\right)+2500$=1000

$x.50=1000-2500$

$x.50=-1500$

$x=-1500:50=-30$

Đào Ngọc Lâm · Answer

(x+1)+(x+3)+(x+5)+...+(x+99)=1000

(�+�+�+...+�)+(1+3+5+7+9+...+99)(x+x+x+...+x)+(1+3+5+7+9+...+99)

(�.50)+2500(x.50)+2500=1000

�.50=1000−2500x.50=1000−2500

�.50=−1500x.50=−1500

�=−1500:50=−30x=−1500:50=−30Câu trả lời: (x+1)+(x+3)+(x+5)+...+(x+99)=1000

(�+�+�+...+�)+(1+3+5+7+9+...+99)(x+x+x+...+x)+(1+3+5+7+9+...+99)

(�.50)+2500(x.50)+2500=1000

�.50=1000−2500x.50=1000−2500

�.50=−1500x.50=−1500

�=−1500:50=−30x=−1500:50=−30