Phân tích đa thức thành nhân tử : (x2 + 5x – 3)(x2 + 5x – 5) – 15
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-x2 - 5x + 24
= -x2 + 3x - 8x + 24
= -x(x + 3) - 8(x - 3)
= (-x - 8)(x + 3)
\(\left(x^2-5x\right)^2-3x^2+15x-18\)
\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+3\right)\)
\(=\left(x^2-5x+3\right)\left(x-6\right)\left(x+1\right)\)
\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)^2-6\left(x^2-5x\right)+3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)\left(x^2-5x-6\right)+3\left(x^2-5x-6\right)\\ =\left(x^2-5x+3\right)\left(x^2-5x-6\right)\\ =\left(x-6\right)\left(x+1\right)\left(x^2-5x+3\right)\)
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)
\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)\\ =\left(x^2+6x-1\right)^2-16-20\\ =\left(x^2+6x-1\right)^2-36\\ =\left(x^2+6x-1-6\right)\left(x^2+6x-1+6\right)\\ =\left(x^2+6x-7\right)\left(x^2+6x+5\right)\\ =\left(x-1\right)\left(x+7\right)\left(x+1\right)\left(x+5\right)\)
\(=\left(x+3\right)^6-y^6\\ =\left[\left(x+3\right)^3-y^3\right]\left[\left(x+3\right)^3+y^3\right]\\ =\left(x+3-y\right)\left[\left(x+3\right)^2+y\left(x+3\right)+y^2\right]\left(x+3+y\right)\left[\left(x+3\right)^2-y\left(x+3\right)+y^2\right]\\ =\left(x+y+3\right)\left(x-y+3\right)\left(x^2+6x+9+xy+3y+y^2\right)\left(x^2+6x+9-xy-3y+y^2\right)\)
\(\left(x^2+6x+9\right)^3-\left(y^2\right)^3=\left(x^2+6x+9-y^2\right)\left[\left(x^2+6x+9\right)^2+\left(x^2+6x+9\right)y^2+y^4\right]\)
\(=\left[\left(x+3\right)^2-y^2\right]\left\{\left[\left(x^2+6x+9\right)^2+2\left(x^2+6x+9\right)y^2+y^4\right]-\left(x^2+6x+9\right)y^2\right\}\)
\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)^2-\left(x+3\right)^2y^2\right]\)
\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)-\left(x+3\right)y\right]\left(x^2+6x+9+y^2\right)+\left(x+3\right)y\)
\(=\left(x+3-y\right)\left(x+3+y\right)\left(x^2+6x+9+y^2-xy-3y\right)\left(x^2+6x+9+y^2+xy+3y\right)\)
(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: \(\left(x^2+1\right)^2+4x\left(x^2-1\right)\)
\(=x^4+2x^2+1+4x^3-4x\)
\(=x^4+2x^3+2x^3+4x^2-2x^2-4x+1\)
\(=\left(x+2\right)\left(x^3+2x^2-2x\right)+1\)
\(\left(x^2-2x-6\right)\left(x^2-2x-11\right)+6\)
\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+66+6\)
\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+72\)
\(=\left(x^2-2x-8\right)\left(x^2-2x-9\right)\)
\(=\left(x-4\right)\left(x+2\right)\left(x^2-2x-9\right)\)
\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15=\left(x^2+5x-3\right)\left(x^2+5x-3-2\right)-15=\left(x^2+5x-3\right)^2-2\left(x^2+5x-3\right)+1-16=\left(x^2+5x-3-1\right)^2-4^2=\left(x^2+5x-4\right)^2-4^2=\left(x^2+5x-8\right)\left(x^2+5x\right)=x\left(x+5\right)\left(x^2+5x-8\right)\)
\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15\)
\(=\left(x^2+5x\right)^2-8\left(x^2+5x\right)-15\)
\(=x\left(x+5\right)\left(x^2+5x-8\right)\)