phân tích đa thức thành nhân tử:
a, x2+5+6
b, x2-4x+3
c, x2+5x+4
d, x2-x-6
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\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15=\left(x^2+5x-3\right)\left(x^2+5x-3-2\right)-15=\left(x^2+5x-3\right)^2-2\left(x^2+5x-3\right)+1-16=\left(x^2+5x-3-1\right)^2-4^2=\left(x^2+5x-4\right)^2-4^2=\left(x^2+5x-8\right)\left(x^2+5x\right)=x\left(x+5\right)\left(x^2+5x-8\right)\)
\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15\)
\(=\left(x^2+5x\right)^2-8\left(x^2+5x\right)-15\)
\(=x\left(x+5\right)\left(x^2+5x-8\right)\)
(1 + x2)2 - 4x(1 - x2)
= (1 + x2)(1 + x2) - 4x(1 - x2)
= (1 + x2 - 4x)(1 + x2 - 1 + x2)
= 2x2(x2 - 4x + 1)
Ta có: \(\left(x^2+1\right)^2+4x\left(x^2-1\right)\)
\(=x^4+2x^2+1+4x^3-4x\)
\(=x^4+2x^3+2x^3+4x^2-2x^2-4x+1\)
\(=\left(x+2\right)\left(x^3+2x^2-2x\right)+1\)
\(=\left(x^2+5x+8\right)\left(x^2+4x+2x+8\right)=\left(x^2+5x+8\right)\left[x\left(x+4\right)+2\left(x+4\right)\right]\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)
\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)
\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)
a) x2+x-2
= x2-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x2+5x+3
= 2x2+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x2+5x-2
= 3x2+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)
\(\left(x^2-2x-6\right)\left(x^2-2x-11\right)+6\)
\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+66+6\)
\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+72\)
\(=\left(x^2-2x-8\right)\left(x^2-2x-9\right)\)
\(=\left(x-4\right)\left(x+2\right)\left(x^2-2x-9\right)\)
-x2 - 5x + 24
= -x2 + 3x - 8x + 24
= -x(x + 3) - 8(x - 3)
= (-x - 8)(x + 3)
Ta có: (x2+6x-5)(x2+6x+3)-20
= [(x2+6x-1)-4][(x2+6x-1)+4]-20
= (x2+6x-1)2-16-20
= (x2+6x-1)2-36
= (x2+6x-7)(x2+6x-5)
= (x+7)(x-1)(x2+6x-5)
\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)\\ =\left(x^2+6x-1\right)^2-16-20\\ =\left(x^2+6x-1\right)^2-36\\ =\left(x^2+6x-1-6\right)\left(x^2+6x-1+6\right)\\ =\left(x^2+6x-7\right)\left(x^2+6x+5\right)\\ =\left(x-1\right)\left(x+7\right)\left(x+1\right)\left(x+5\right)\)
\(\left(x^2-5x\right)^2-3x^2+15x-18\)
\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+3\right)\)
\(=\left(x^2-5x+3\right)\left(x-6\right)\left(x+1\right)\)
\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)^2-6\left(x^2-5x\right)+3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)\left(x^2-5x-6\right)+3\left(x^2-5x-6\right)\\ =\left(x^2-5x+3\right)\left(x^2-5x-6\right)\\ =\left(x-6\right)\left(x+1\right)\left(x^2-5x+3\right)\)
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
b) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-3\right)\left(x-1\right)\)
c) \(x^2+5x+4=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+4\right)\left(x+1\right)\)
d) \(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
cảm ơn nha