Question

Dạng 4: Phương pháp thêm, bớt một hạng tử.

Bài 1; Phân tích đa thức thành nhân tử

a) x^4 + 16;

b) x^4y^4 + 64;

c) x^4y^4 + 4;

f) x^8 + x + 1;

g) x^8 + x^7 + 1;

h) x^8 + 3x^4 + 1;

k) x^4 + 4y^4;

l) 4x^4 + 1;

Bài 2: phân tích đa thức thành nhân tử

a) a^2 - b^2 - 2x(a-b) ;

b) a^2 - b^2 - 2x(a+b);

Answer 1

\(x^4y^4+64=x^4y^4+16x^2y^2+64-16x^2y^2=\left(x^2y^2+8\right)^2-16x^2y^2=\left(x^2y^2-4xy+8\right)\left(x^2y^2+4xy+8\right)\)

\(x^8+x+1=x^8-x^2+\left(x^2+x+1\right)=x^2\left(x^6-1\right)+\left(x^2+x+1\right)=x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)+x^2+x+1=\left(x^2+x+1\right)\text{[}x^2\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)+1\text{]}\)

\(g,tach:x^2+x+1\)

\(x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=\left(x^2+2y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)

Answer 2

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

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