S = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷

= (1 + 2) + (2² + 2³) + (2⁴ + 2⁵) + (2⁶ + 2⁷)

= (1 + 2) + 2²(1 + 2) + 2⁴(1 + 2) + 2⁶(1 + 2)

= (1 + 2)(1 + 2² + 2⁴ + 2⁶) 

= 3(1 + 2² + 2⁴ + 2⁶) \(⋮3\)(ĐPCM)

2S = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ 

S = (1+2 ) + (2² + 2³ ) + (2⁴ + 2⁵ ) + (2⁶ +2⁷)

S = 3 + 2²(1+2) + 2⁴(1+2) + 2⁶(1+2)

S = 3+2².3 + 2⁴.3 + 2⁶ .3 

S = 3(1+2² + 2⁴ + 2⁶ ) \(⋮\) 3

=> đpcm

Vì 1+2=3 \(⋮\)3 =>(1+2)+2²+2³+2⁴+2⁵+2⁶+2⁷=>S \(⋮\)3

Vậy S \(⋮\)9

Nhớ k cho mình nhé !

\(\frac{-3.7^4+7^3}{7^5.6-7^3.2}=\frac{7^3\left(-3.7+1\right)}{7^3\left(7^2.6-2\right)}=\frac{-21+1}{49.6-2=4}=\frac{-20}{290}\)

A=2+2²+2³+2⁴+2⁵+2⁶

=2.(1+2+2²+2³+2⁴+2⁵)

=2.(1+2+4+8+16+32)

=2.63

=2.7.9 chia hết cho 9(vì trong tích có 1 thừa số là 9)

Chúc bạn học giỏi nha!!!

K cho mik vs nhé Mai Phước Trí

    (-5).7.(-12).|-4|.(-8)

= -35.(-12).4.(-8)

=420.4.(-8)

=1680.(-8)

=-13440

1/ (-5).7.(-12).|-4|.(-8) = -13440

2/ x-(-19)+(-2)mũ 5=14-(-2)mũ 4 => x=(-7)

3/ x+4y-xy =15; không có x;y thỏa mãn đề bài, nếu 3x+4y-xy =16 thì x=8;y=2

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!

a: m>n

=>2m>2n

=>2m-2>2n-2

b: m>n

=>-3m<-3n

=>-3m+1<-3n+1

c: m>n

=>2m>2n

=>2m+3>2n+3

mà 2n+3>2n+1

nên 2m+3>2n+1

d: m>n

=>-5m<-5n

=>-5m+3<-5n+3

mà -5n+3<-5n+7

nên -5m+3<-5n+7