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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
\(1,3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3[\left(x^2-2xy+y^2\right)-4z^2]\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
\(=3\left(x-y-2x\right)\left(x-y+2z\right)\)
\(2,x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=\left(x^2-5x\right)+\left(3x-15\right)\)
\(=x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x+3\right)\left(x-5\right)\)
\(x^6-9x^5+30x^4-45x^3+30x^2-9x+1\)
\(=\left(x^2\right)^3-9x^5+30x^4-45x^3+30x^2-9x+1^3\)
\(=\left(x^3-3x+1\right)^3\)
Rút gọn
\(\left(2x+1\right)\left(4x^2-3x+1\right)+\left(2x-1\right)\left(4x^2+3x+1\right)\)
\(=8x^3-12x^2+2x+4x^2-3x+1+8x^3+12x^2+2x-4x^2-3x-1\)
\(=16x^3-2x\)
Phân tích đa thức thnahf nhân tử
\(4y^2+16y-x^2-8x\)
\(=\left(4y^2-x^2\right)+\left(16y-8x\right)\)
\(=\left(2y-x\right)\left(2y+x\right)+8\left(2y-x\right)\)
\(=\left(2y-x\right)\left(2y+x+8\right)\)
Chứng minh .............
Có: \(x^2+x+1=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Kết luận......
1: \(x^2-2x-24=\left(x-6\right)\left(x+4\right)\)
2: \(x^2-8x+15=\left(x-3\right)\left(x-5\right)\)
3: \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)