Câu hỏi của Trần Minh Ánh

Question

Tìm số nguyên x:

x+(x+1)+(x+2)+(x+3)+....+(x+13)+(x+14)=14

%Hz@ · Accepted Answer

$x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14$

$x+x+1+x+2+x+3+...+x+14=14$

$15x+\left(1+2+3+...+14\right)=14$

$15x+\frac{\left(14+1\right).14}{2}=14$

$15x+105=14$

$15x=-91$

$x=-\frac{91}{15}=-6,06$

Câu trả lời:

$x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14$

$x+x+1+x+2+x+3+...+x+14=14$

$15x+\left(1+2+3+...+14\right)=14$

$15x+\frac{\left(14+1\right).14}{2}=14$

$15x+105=14$

$15x=-91$

$x=-\frac{91}{15}=-6,06$

Hoàng hôn ( Cool Team ) · Answer

Ko chắc lắm

x+(x+1)+(x+2)+(x+3)+...+(x+13)+(x+14)=14

x+x+1+x+2+x+3+...+x+13+x+14=14

(x+x+x+x+...+x+x)+(1+2+3+...+14)=14

15x + 14 = 14

15x = 14 - 14

15x = 0

x = 0:15

x = 0

Vậy x=0