Ta có: \(7^{98}+7^{99}+7^{100}=7^2\left(7^{98}+7^{97}+7^{96}\right)⋮\left(7^{98}+7^{97}+7^{96}\right)\)

Vậy \(\left(7^{98}+7^{99}+7^{100}\right)⋮\left(7^{98}+7^{97}+7^{96}\right)\left(đpcm\right)\)

Ta có: \(7^{98}+7^{99}+7^{100}=7^2\left(7^{98}+7^{97}+7^{96}\right)⋮\left(7^{98}+7^{97}+7^{96}\right)\)

Vậy \(\left(7^{98}+7^{99}+7^{100}\right)⋮\left(7^{98}+7^{97}+7^{96}\right)\)

\(\rightarrowđpcm\)

1)-73

2)-5

3)-64

4)-56

5)9

6)-20

7)6

8)-2

Tick nha

mk ra kết quả thôi nha

/ là phần

tui làm sai rồiiiiiiiiiiiii

= 0 nhé'

quá dễ, dễ quá

ko chia đc vì cái cuối =0 ko thể chia 0

lua nguoi

a)=7x(1/4+1/28+1/70+1/130)

  =7x(1/1x4+1/4x7+1/7x10+1/10x13)

  =7x(1-1/4+1/4-1/7+1/7-1/10+1/10-1/13)

  =7x(1+1/13)

  =7x14/13=98/13

Các câu còn lại p dựa theo cách trên làm nốt nhé!!!

Lời giải chi tiết:

\(\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}\)= \(\frac{15-15\left(\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}{18-18\left(\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}\)=\(\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)= \(\frac{15}{18}\)=\(\frac{5}{6}\)

cảm ơn nhìu

\(\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}=\frac{15\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}{18\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)