Dùng phương pháp bổ sung số thích hợp để được hằng đẳng thức :
1) 6x2 + 7x - 3
2) 5x2 + 14x - 3
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Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
\(1,2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(2x+1\right)\left(x-2\right)\)
\(2,4x^2-7x-2\)
\(=4x^2-8x+x-2\)
\(=4x\left(x-2\right)+x-2\)
\(\left(4x+1\right)\left(x-2\right)\)
\(2x^2-7x+3\)
\(=2\left(x^2-\frac{7}{2}x+\frac{3}{2}\right)\)
Vậy thôi đâu cần dùng HĐT
c ) \(x^2-7x+12\)
\(=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
d ) \(x^2+7x+12\)
\(=\left(x^2+3x\right)+\left(4x+12\right)\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
4x^2 - 7x -2 = 4x^2 - 8x + x - 2 = 4x(x - 2) + (x - 2) = (x -2)(4x + 1)
\(1)6x^2+7x-3\\=6x^2+9x-2x-3\\=3x(2x+3)-(2x+3)\\=(2x+3)(3x-1)\\2)5x^2+14x-3\\=5x^2+15x-x-3\\=5x(x+3)-(x+3)\\=(x+3)(5x-1)\)
1) \(6x^2+7x-3=6x^2+9x-2x-3=3x\left(x+3\right)-\left(2x+3\right)=\left(2x+3\right).\left(3x-1\right)\)
2) \(5x^2+14x-3=5x^2+15x-x-3=5x\left(x+3\right)-\left(x+3\right)=\left(x+3\right).\left(5x-1\right)\)