Giải thích các bước giải:
c.Ta có:
$C=x^4-x^3y+2x^3+x^2-xy+4x-2y+5$
$\to C=(x^4-x^3y)+2x^3+(x^2-xy)+2x+(2x-2y)+5$
$\to C=x^3(x-y)+2x^3+x(x-y)+2x+2(x-y)+5$
Mà $x-y=-2$
$\to C=-2x^3+2x^3-2x+2x-4+5$
$\to C=1$
b.Sửa $-8x^0$ thành $+8x^0$ cho đúng quy luật
Ta có:
$B=x^{100}-8x^{98}+8x^{96}-...-8x^2+8x^0$
$\to x^2B=x^{102}-8x^{100}+8x^{98}-...-8x^4+8x^2$
$\to x^2B+B=x^{102}-7x^{100}+8x^0$
$\to (x^2+1)B=x^{102}-7x^{100}+8$
Mà $x=3$
$\to (3^2+1)B=3^{102}-7\cdot 3^{100}+8$
$\to 10B=3^{102}-7\cdot 3^{100}+8$
$\to B=\dfrac{3^{102}-7\cdot 3^{100}+8}{10}$
$\to B=\dfrac{3^2\cdot 3^{100}-7\cdot 3^{100}+8}{10}$
$\to B=\dfrac{9\cdot 3^{100}-7\cdot 3^{100}+8}{10}$
$\to B=\dfrac{2\cdot 3^{100}+8}{10}$
