Bài 2:
a: \(25^{n+1}-25^n=25^n\left(25-1\right)=25^n\cdot24\)
\(=4\cdot6\cdot25\cdot25^{n-1}=100\cdot6\cdot25^{n-1}⋮100\)
b: \(B=50^{n+2}-50^{n+1}\)
\(=50^{n+1}\left(50-1\right)\)
\(=50^n\cdot50\cdot49=50^n\cdot10\cdot245⋮245\)
c: \(n^3-n=n\left(n^2-1\right)=n\cdot\left(n-1\right)\left(n+1\right)\)
Vì n-1;n;n+1 là ba số nguyên liên tiếp
nên \(n\left(n-1\right)\left(n+1\right)⋮3!=6\)
=>\(n^3-n⋮6\)
Bài 3:
a: \(D=4\left(x-2\right)\left(x+1\right)+\left(2x-4\right)^2+\left(x+1\right)^2\)
\(=4\left(x^2-x-2\right)+4x^2-16x+16+x^2+2x+1\)
\(=4x^2-4x-8+5x^2-14x+17\)
\(=9x^2-18x+9=9\left(x^2-2x+1\right)=9\left(x-1\right)^2\)
Thay x=-1/2 vào D, ta được:
\(D=9\left(-\dfrac{1}{2}-1\right)^2=9\cdot\left(-\dfrac{3}{2}\right)^2=9\cdot\dfrac{9}{4}=\dfrac{81}{4}\)
b: \(E=x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2z-xy^2+xz^2-z^2y\)
\(=x^2y-xy^2-x^2z+y^2z+xz^2-z^2y\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Thay x=6;y=5;z=4 vào E, ta được:
E=(6-5)(6-4)(5-4)=2
c: x=9 nên x+1=10
\(E=x^{2017}-10x^{2016}+10x^{2015}-...-10x^2+10x-10\)
\(=x^{2017}-x^{2016}\cdot\left(x+1\right)+x^{2015}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-\left(x+1\right)\)
\(=x^{2017}-x^{2017}-x^{2016}+x^{2016}+...+x^2+x-x-1\)
=-1
