Bài 1:phân tích đa thức thành nhân tử
A=x2+6x+5
B=(x2-x+1).(x2-x+2)-12
C=x2.(y2-4)2-6x(y2-4)+9
Bài 2:phân tích đa thức thành nhân tử
A=a4+a3+a3b+a2b
B=a3+4a2+4a+3
C=a3+3a2+4a+3
D=x2y+xy2x2z+xz2+y2z+yz2+2xy2
Bài 3:phân tích đa thức thành nhân tử
A=x3+9x2+26x+24
B=x5+x4+1
c=x4-4x3+8x+3
Các bạn giúp mình với ạ. Mình bí quá
\(A=a^4+a^3+a^3b+a^2b\)
\(A=a\left(a^3+a^2\right)+b\left(a^3+a^2\right)\)
\(A=\left(a+b\right)\left(a^3+a^2\right)\)
\(A=a^2\left(a+1\right)\left(a+b\right)\)
Bài 1:
\(A=x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\)
Đặt \(a=x^2-x+2\) ta có:
\(B=\left(a-1\right).a-12=a^2-a-12=a^2+3a-4a-12=a\left(a+3\right)-4\left(a+3\right)=\left(a+3\right)\left(a-4\right)\)
Thay a = x2 - x + 2 vào ta được:
\(\left(x^2-x+2-4\right)\left(x^2-x+2+3\right)=\left(x^2-x-2\right)\left(x^2-x+5\right)=\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)\)
Bài 2:\(B=a^3+4a^2+4a+3=a^3+3a^2+a^2+3a+a+3=a^2\left(a+3\right)+a\left(a+3\right)+\left(a+3\right)=\left(a+3\right)\left(a^2+a+1\right)\)
Bài 3:\(A=x^3+9x^2+26x+24=x^3+2x^2+7x^2+14x+12x+24=x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)=\left(x+2\right)\left(x^2+7x+12\right)=\left(x+2\right)\left(x^2+7x+12\right)=\left(x+2\right)\left[x^2+3x+4x+12\right]=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(B=x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
A=x\(^2\)+6x+5
A=x\(^2\)+x+5x+5
A=(x\(^2\)+x)+(5x+5)
A=x(x+1)+5(x+1)
A=(x+1)(x+5)
Bài 1:
A=x2+6x+5=x2+5x+x+5=x(x+5)+(x+5)=(x+1)(x+5)A=x2+6x+5=x2+5x+x+5=x(x+5)+(x+5)=(x+1)(x+5)
Đặt a=x2−x+2a=x2−x+2 ta có:
B=(a−1).a−12=a2−a−12=a2+3a−4a−12=a(a+3)−4(a+3)=(a+3)(a−4)B=(a−1).a−12=a2−a−12=a2+3a−4a−12=a(a+3)−4(a+3)=(a+3)(a−4)
Thay a = x2 - x + 2 vào ta được:
(x2−x+2−4)(x2−x+2+3)=(x2−x−2)(x2−x+5)=(x+1)(x−2)(x2−x+5)
