Phân tích a^3-8+6a^2-12a thành nhân tử

1)\(a^3-8+6a^2-12a=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)=\left(a-2\right)\left(a^2+8a+4\right)\)

2)\(\left(a+b\right)^3+\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3=2a^3+6ab^2=2a\left(a^2+3b^2\right)\)

Câu 5: Phân tích đa thức thành nhân tử.

a) x²-7x+6. b)x⁴+64. c)x²+12x+35. d) 4x⁴+1

e)x²-x-56. f)4x⁴+81. g) 5x²-x-4. h) 64x⁴+y⁴

\(a,x^2-7x+6=x^2-6x-x+6\)

\(=x\left(x-6\right)-\left(x-6\right)=\left(x-1\right)\left(x-6\right)\)

\(b,x^4+64=\left(x^4+16x^2+64\right)-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

\(c,x^2+12x+35=x^2+5x+7x+35\)

\(=x\left(x+5\right)+7\left(x+5\right)=\left(x+7\right)\left(x+5\right)\)

\(d,4x^4+1=\left(4x^4+4x^2+1\right)-4x^2\)

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

\(e,x^2-x-56=x^2-8x+7x-56\)

\(=x\left(x-8\right)+7\left(x-8\right)=\left(x+7\right)\left(x-8\right)\)

\(f,4x^2+81=\left(4x^4+36x^2+81\right)-36x^2\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2=\left(2x^2+6x+9\right)\left(2x^2-6x+9\right)\)

\(g,5x^2-x-4=5x^2-5x+4x-4\)

\(=5x\left(x-1\right)+4\left(x-1\right)=\left(5x+4\right)\left(x-1\right)\)

\(h,64x^4+y^4=\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\)

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

Phân tích đa thức thành nhân tử:

x⁴+2x³+2x²+2x¹

\(x^4+2x^3+2x^2+2x=x\left(x^3+2x^2+2x+2\right)\)

Bài 1: phân tích đa thức sau thành nhân tử( thêm bớt cùng 1 hạng tử)

x^3+x^2+4

\(x^3+x^2+4\)

\(=x^3+x^2+4+4x-4x\)

\(=x^3-4x+x^2+4+4x\)

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

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

\(=\left(x+2\right)\left[x\left(x-2\right)+\left(x+2\right)\right]\)

\(=\left(x+2\right)\left[x^2-2x+x+2\right]\)

\(=\left(x+2\right)\left[x^2-x+2\right]\)

\(=\left(x+2\right)\left[\left(x^2-x+0,25\right)+0,75\right]\)

\(=\left(x+2\right)\left[\left(x-0,5\right)^2+0,75\right]\)

Phân tích các đa thức sau thành nhân tử:

\(1,y^2+8y+7\)

\(2.6x^2-7x-20\)

\(3.x^2-6x-y^2-4y+5\)

4.\(4.x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

1. y²+8y+7=y²+y+7y+7=y(y+1)+7(y+1)=(y+7)(y+1)

2. 6x²-7x-20=6x²-15x+8x-20=3x(2x-5)+4(2x-5)=(3x+4)(2x-5)

Phân tích thành nhân tử:

a) 16\(x^2\)- 4\(y^2\)

b) mx - my - nx - ny + \(y^2\)- 2xy + \(x^2\)

c) \(\left(x+y+z\right)^3\)- \(x^3\)- \(y^3\)- \(z^3\)

d) \(x^2\) - 7xy + 10\(y^2\)

e) \(x^2\)+ x -6

g) \(x^4\) + \(x^2\)+2x

\(A=x^4+4\)

\(A=x^4+4+4x^2-4x^2\)

\(A=\left(x^2+2\right)^2-4x^2\)

\(A=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)

\(B=x^4+64\)

\(B=x^4+64+16x^2-16x^2\)

\(B=\left(x^2+8\right)^2-16x^2\)

\(B=\left(x^2+8+4x\right)\left(x^2+8-4x\right)\)

Phân tích thành nhân tử?
x^4+4
x^4 + 64
( Giúp mình nhanh với nhé ạ, mình gấp lắm, cảm ơn các bạn )

Phân tích đa thức sau thành nhân tử:

\(x^4+x^2+1\)

phân tích đa thức thành phân tử:

x³+9x²+26x+24

\(a,16x^2-4y^2\)

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

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

\(=\left[2\left(2x-y\right)\right]\left[2\left(2x+y\right)\right]\)

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

\(b,mx-my-nx+ny+y^2-2xy+x^2\)

\(=\left(mx-my\right)-\left(nx-ny\right)+\left(y^2-2xy+x^2\right)\)

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

\(=\left(x-y\right)\left(m-n-x+y\right)\)

\(c,\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y+z\right)^3-x^3\right]-\left[y^3+z^3\right]\)

\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+\left(x+y+z\right)x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left[\left(x+y+z\right)^2+x^2+xy+z^2-y^2+yz-z^2\right]\)

\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz+x^2+xy+z^2-y^2+yz-z^2\right)\)

\(=\left(y+z\right)\left(2x^2+z^2+3xy+3yz+2xz\right)\)

\(x^3+9x^2+26x+24\)

\(=x^3+4x^2+5x^2+20x+6x+24\)

\(=x^2\left(x+4\right)+5x\left(x+4\right)+6\left(x+4\right)\)

\(=\left(x^2+5x+6\right)\left(x+4\right)\)

\(x^4 + x^2 + 1 \)

\(= (x^4 + 2x^2 + 1) - x^2 \)

\(= (x^2 + 1)^2 - x^2 \)

\(= (x^2 + 1 + x).(x^2 + 1 - x) \)