Question

3) tìm x

a)  (x+1) +(x+2)+(x+3)+(....)+(x+99)=90387

b)  (1+4+7+.....+100)

c)  1+2+3+......+x=999

 

Answer 1

a, \(\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=90387\)

\(< =>99x+\left(1+2+3+...+99\right)=90387\)

\(< =>99x+\left(\frac{100.99}{2}\right)=90387\)

\(< =>99x=90387-4950=85437\)

\(< =>x=\frac{85437}{99}=863\)

Answer 2

b,\(1+4+7+...+100\)

Số số hạng : \(\left(100-1\right):3+1=34\)

Tổng \(\frac{\left(100+1\right).34}{2}=1717\)

Vậy \(1+4+7+...+100=1717\)

c, \(1+2+3+...+x=999\)

\(< =>\frac{\left(x+1\right)x}{2}=999\)

\(< =>x^2+x-1998=0\)

\(< =>\orbr{\begin{cases}x=\frac{-1+\sqrt{7993}}{2}\\x=\frac{-1-\sqrt{7993}}{2}\end{cases}}\)