phần b là 3x+2y, còn phần a là 3x-4y nhé

Ta có: x = 9 => x - 9 = 0

\(Q\left(x\right)=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-9x^{13}-x^{13}+9x^{12}+x^{12}-9x^{11}+...-x^3+9x^2+x^2-9x-x+9+1\)

\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+...-x^2\left(x-9\right)+x\left(x-9\right)-\left(x-9\right)+1\)

\(=0+1=1\)

\(A=6xy\left(xy-y^2\right)-8x^2.\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(A=6x^2y^2-6xy^3-8x^3+8x^2y^2+5y^2x^2-5xy^3\)

\(A=19x^2y^2-11xy^3-8x^3\)

Tại x=1/2, y=2

\(A=19.\frac{1}{4}.2^2-11.\frac{1}{2}.2^3-8\left(\frac{1}{2}\right)^3=19-44-1=-26\)

Q(x) có thể làm như sau:

Vì x=9 nên x+1 = 10

Thay 10=x+1 ta có

\(Q\left(x\right)=x^{14}-\left(x+1\right).x^{13}+\left(x+1\right).x^{12}-\left(x+1\right).x^{11}+...+\left(x+1\right).x^2-\left(x+1\right)x+x+1\)

\(Q\left(x\right)=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^2-x^2-x+x+1\)

Q(x) = 1

c: \(x^2-10x+21=\left(x-3\right)\left(x-7\right)\)

a: \(x^2y+xy^3-xy-y^3\)

\(=xy\left(x-1\right)+y^3\left(x-1\right)\)

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

\(a) x^2y+xy^3-xy-y^3\\=(x^2y+xy^3)-(xy+y^3)\\=xy(x+y^2)-y(x+y^2)\\=(x+y^2)(xy-y)\\=y(x+y^2)(x-1)\\b)2x^2+5x+8(xem lại đề)\\c)x^2-10x+21\\=x^2-3x-7x+21\\=x(x-3)-7(x-3)\\=(x-3)(x-7)\)

\(a,=xy\left(x+y^2\right)-y\left(x+y^2\right)=y\left(x+y^2\right)\left(x-1\right)\\ c,=x^2-7x-3x+21=\left(x-7\right)\left(x-3\right)\)

a, \(x^2y+xy^2-xy-y^3=y\left(x^2+xy-x-y^2\right)\)

c, \(x^2-10x+21=\left(x-3\right)\left(x-7\right)\)

a) \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)

b) \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)\left(x+1\right)\left(y-z\right)\)

c) \(15x+15y-x^2-xy=15\left(x+y\right)-x\left(x+y\right)=\left(15-x\right)\left(x+y\right)\)

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

khó .mình chịu

Hình như bài này lớp 6 cx có

k cho mk nha