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\(a,\)\(x^3-13x-12\)
\(=x^3-x-12x-12\)
\(=x\left(x^2-1\right)-12\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
a) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen
c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha
\(x^4-2x^3+2x-1=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(x^3-x^2-x+1\right)=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]=\left(x-1\right)^2\left(x^2-1\right)=\left(x-1\right)^3\left(x+1\right)\)
\(x^4-2x^3+2x-1\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)\)
\(6x^2-13x+6\)
\(=6x^2-9x-4x+6\)
\(=\left(2x-3\right)\left(3x-2\right)\)
\(=6x^2+9x+4x+6\)
\(=3x.\left(2x+3\right)+2.\left(2x+3\right)\)
\(=\left(2x+3\right).\left(3x+2\right)\)
\(\)
\(\left(x^2-2x-6\right)\left(x^2-2x-11\right)+6\)
\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+66+6\)
\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+72\)
\(=\left(x^2-2x-8\right)\left(x^2-2x-9\right)\)
\(=\left(x-4\right)\left(x+2\right)\left(x^2-2x-9\right)\)
\(=x^2\left(x^2+2x+1\right)+x+1\)
\(=x^2\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(x^4+2x^3+x^2+x+1\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
1. Phân tích đa thức thành nhân tử
a, 1/4x^2-5xy+25y^2
b, (7x-4)^2-(2x+1)^2
c, (x-2)^2-4y
d, 125-x^6
a) \(\frac{1}{4}x^2-5xy+25y^2=\left(\frac{1}{2}x\right)^2-5xy+\left(5y\right)^2\)
\(=\left(\frac{1}{2}x-5y\right)^2\)
b) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4+2x+1\right)\times\left(7x-4-2x-1\right)=\left(9x-3\right)\times\left(5x-5\right)\)
\(=3\times5\times\left(3x-1\right)\times\left(x-1\right)=15\times\left(3x-1\right)\times\left(x-1\right)\)
c)\(\left(x-2\right)^2-4y^2=\left(x-2-2y\right)\left(x-2+2y\right)\)
d) \(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)
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