5⁵-5⁴+5³=5³.(5²-5¹+5⁰)=5³.(25-5+1)=5³.21=5³.3.7 chia hết cho 7

=>ĐPCM

7⁶+7⁵-7⁴=7⁴.(7²+7¹-7⁰)=7⁴.(49+7-1)=7⁴.55=7⁴.5.11 chia hết cho 11

=>ĐPCM

a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)

b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)

c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)

d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)

\(S=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{11}\left(1+2\right)=3\left(2+2^3+...+2^{11}\right)⋮3\)

\(S=2\left(1+2+2^2\right)+...+2^{10}\left(1+2+2^2\right)=7\left(2+...+2^{10}\right)⋮7\)

Vì S chia hết cho 2 và S chia hết cho 3 

nên \(S⋮6\)

1) 5⁵ - 5⁴ + 5³ = 5³ . 5² - 5³ . 5 - 5³

= 5³ . ( 5² - 5 + 1 )

= 5³ . ( 25 - 5 - 1 )

= 5³ . 21

= 5³ . 3 . 7 chia hết cho 7

Vậy chứng minh 5⁵ - 5⁴ + 5³ chia hết cho7

2) 7⁶ + 7⁵ - 7⁴ = 7⁴ . 7² + 7⁴ . 7 - 7⁴

= 7⁴ . ( 7² + 7 - 1 )

= 7⁴ . ( 49 + 7 - 1 )

= 7⁴ . 55

= 7⁴ . 5 .11 chia hết cho 11

Vậy chứng minh 7⁶ + 7⁵ - 7⁴ chia hết cho 11

Tích mình nha !!!!!!!!!!!!!!!!!

a: \(=\dfrac{4}{9}\left(4+\dfrac{1}{3}+13+\dfrac{2}{3}\right)=\dfrac{4}{9}\cdot18=8\)

b: \(=\dfrac{3}{8}\left(5+\dfrac{1}{4}+10+\dfrac{3}{4}\right)=\dfrac{3}{8}\cdot16=6\)

c: \(=\dfrac{-2}{7}\left(6+\dfrac{1}{5}+14+\dfrac{4}{5}\right)=\dfrac{-2}{7}\cdot21=-6\)

5) x^{m + 2} - x^{m + 1}

= x^{m + 1} (x - 1)

1: \(=5x^2y\left(x-7\right)+5xy\left(x-7\right)\)

\(=5xy\left(x-7\right)\left(x+1\right)\)

2: \(=3ab\left(x-y\right)-3a\left(x-y\right)\)

\(=3a\left(x-y\right)\left(b-1\right)\)

3: \(=4a\left(x-5\right)+2\left(x-5\right)\)

\(=2\left(x-5\right)\left(2a+1\right)\)

\(\overline{aaa}+\overline{bbb}=111.a+111.b=111\left(a+b\right)=37\times3\times\left(a+b\right)⋮37\)

KHÔNG ĐÚNG : 

NẾU \(a+b=c+b\Rightarrow a=c\)

Thì mơi đúng như trên không phải

MK GHI THIỂU NHA TH TRÊN LÀ TH ĐẶC BIỆT VÌ =0 CÒN NẾU KHÔNG BẰNG KHÔNG THÌ GIẢI NHƯ VẬY LÀ SAI 

\(a,10.a^6+20a^5=10a^5\left(a+2\right)\)

\(b,5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

\(c,3ab^3+6ab^2-18ab=3ab\left(b^2+2b-1\right)\)

\(d,15x^3y^2+10x^2y^2-20x^2y^3=5x^2y^2\left(3x+2-4y\right)\)

\(e,a^2\left(x-1\right)-b\left(1-x\right)=a^2\left(x-1\right)+b\left(x-1\right)=\left(x-1\right)\left(a^2+b\right)\)

\(f,x\left(x-5\right)-4\left(5-x\right)=x\left(x-5\right)+4\left(x-5\right)=\left(x-5\right)\left(x+4\right)\)

(mk sửa lại thứ tự là a,b,c,d,e,f nha)

chúc bn học tốt

\(1,10a^6+20a^5=10a^5\left(a+10\right)\)

\(2,5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)\)

\(=5\left(x-y\right)^2\)

\(3,3ab^3+6ab^2-18ab\)

\(=3ab\left(b^2+2b-6\right)\)

\(4,15x^3y^2+10x^2y^2-20x^2y^3\)

\(=5x^2y^2\left(3x+2-4y\right)\)

\(5,a^2\left(x-1\right)-b\left(1-x\right)\)

\(=a^2\left(x-1\right)+b\left(x-1\right)\)

\(=\left(x-1\right)\left(a^2+b\right)\)

\(6,x\left(x-5\right)-4\left(5-x\right)\)

\(=x\left(x-5\right)+4\left(x-5\right)\)

\(=\left(x+4\right)\left(x-5\right)\)

mk sửa lại câu 3 chút nha
\(3ab^3+6ab^2-18ab=3ab\left(b^2+2b-6\right)\)

xl vì sai sót
chúc bn học tốt nhé