\(x^3-x^2-2x^2+2x\)

\(=x^2\left(x-1\right)-2x\left(x-1\right)\)

\(=\left(x^2-2x\right)\left(x-1\right)\)

\(=\left(x-1\right)\left(x-2\right)x\)

Vì đây là tích 3 số tự nhiên liên tiếp nên sẽ chia hết cho 6

B = (n^2 - 2n + 1)^3 

= [(n-1)^2]^3

= (n-1)^6 ⋮ (n - 1)^2 

đpcm

\(B=\left(n^2-2n+1\right)^3=\left[\left(n-1\right)^2\right]^3=\left(n-1\right)^6\)

\(B\div\left(n-1\right)^2=\left(n-1\right)^6\div\left(n-1\right)^2=\left(n-1\right)^4\)

=> Đpcm

Ta có:

$A=\frac{3^{10}.11+3^{10}.5}{3^9{.2}^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}$

$=\frac{3^{10}.16}{3^9.16}=\frac{3.1}{1.1}=3$

Vậy giá trị biểu thức A là 3

a: \(a^3-a=a\left(a-1\right)\left(a+1\right)\)

Vì a;a-1;a+1 là ba số nguyên liên tiếp

nên \(a\left(a-1\right)\left(a+1\right)⋮3!\)

hay \(a^3-a⋮6\)

= 3¹⁵( 1 + 3 + 3²)

= 3¹⁵ . 13 : hết cho 13

`= 3^15(1+3+3^2)`

`=3^15 .13`

`=>3^15+3^16+3^17 vdots 13`

\(3^{15}+3^{16}+3^{17}\)

\(=3^{15}\left(1+3+9\right)\)

\(=3^{15}.13\)

\(\Rightarrow3^{15}\times13⋮3\)

Vậy \(3^{15}+3^{16}+3^{17}⋮3\)

a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)

\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)

\(=2^4.5+2-5^2\)

\(=57\)

b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)

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

c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)

\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)

\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)