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(x2+x)2+4x2+4x+12 = (x2+x)2+4(x2+x)+12
= (x2+x) (x2 + x +4)+12
Đặt x2 + 4x + 8 = A. Ta sẽ được:
A2 + 3xA + 2x2
= A2 - xA - 2xA + 2x2
= A(A-x) - 2x(A-x)
= (A-x)(A-2x)
= (x2+3x+8)(x2+2x+8)
ta có
x3-3x2-4x+12
=x2(x-3) -4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
bn k mk nha
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(x^3-4x-12+3x^2=x\left(x^2-2^2\right)+3\left(x^2-2^2\right)=\left(x-2\right)\left(x+2\right)\left(x+3\right)\)
\(x^2+2xy-15y^2=x^2+2xy+y^2-16y^2=\left(x+y\right)^2-\left(4y\right)^2=\left(x-3y\right)\left(x+5y\right)\)
\(\left(x-y\right)^2-6\left(x-y\right)-16=\left(x-y\right)^2-2\times\left(x-y\right)\times3+9-25=\left(x-y-3\right)^2-5^2=\left(x-y-8\right)\left(x-y+2\right)\)
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-4x^2\)
\(=\left(x+2\right)\left(x+12\right)\left(x+3\right)\left(x+8\right)-4x^2\)
\(=\left(x^2+14x+24\right)\left(x^2+11x+24\right)-\left(2x\right)^2\)
Đặt \(x^2+11x+24=a\)
\(=a\left(a+3x\right)-4x^2=a^2+3ax-4x^2=a^2-ax+4ax-4x^2=\left(a-x\right)\left(a+4x\right)\)