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1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)
2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
Ta có: \(1+6x-6x^2-x^3\)
\(=-x^3-6x^2+6x+1\)
\(=\left(-x^3+1\right)-6x\left(x-1\right)\)
\(=-\left(x-1\right)\left(x^2+x+1\right)-6x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\left(-x^2-x-1-6x\right)\)
\(=-\left(x-1\right)\left(x^2+7x+1\right)\)
1. \(=3x\left(2x+5\right)\)
2. \(=\left(3x-1\right)\left(3x+1\right)\)
3. \(=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
1, = 3x.(2x + 5)
2. = (3x)2 - 12 = (3x - 1).(3x +1 )
3, =(x2 + 6x + 9) - y2 = (x + 3)2 - y2 =(x + y -3 ). (x - y +3)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
1) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
\(3x^2+x-2=3x^2-2x+3x-2=x\left(3x-2\right)+\left(3x-2\right)=\left(x+1\right)\left(3x-2\right)\)
\(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(x^2+2xy-15y^2=x^2-3xy+5xy-15y^2=x\left(x-3y\right)+5y\left(x-3y\right)=\left(x+5y\right)\left(x-3y\right)\)
a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)
b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)
d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)
e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)
1)
\(y^2-4y+4-x^2\\ =\left(y-2\right)^2-x^2\\ =\left(y-2-x\right)\left(y-2+x\right)\)
2)
\(8x^3-12x^2+6x-2\\ =2\left(4x^3-6x^2+3x-1\right)\\ =2\left(4x^3-4x^2-2x^2+2x+x-1\right)\\ =2\left(4x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\right)\\ =2\left(x-1\right)\left(4x^2-2x+1\right)\)
1) \(y^2-4y+4-x^2\)
\(=\left(y^2-4y+4\right)-x^2\)
\(=\left(y-2\right)^2-x^2\)
\(=\left(y-2-x\right)\left(y-2+x\right)\)
2) \(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=\left(2x-1\right)^3\)
\(=\left(2x-1\right)\left(2x-1\right)\left(2x-1\right)\)
a) (x + 2)(x + 4). b) 2(x + 6)(x + l).
c) 3(3x + 5)(x + l). d) (6x -7y)(x + y).
mk chỉnh đề
\(x^2-x-2=x^2-2x+x-2=x\left(x-2\right)+\left(x-2\right)=\left(x+1\right)\left(x-2\right)\)
\(x^2+6x+7=x^2+6x+9-2=\left(x+3\right)^2-2=\left(x+3-\sqrt{2}\right)\left(x+3+\sqrt{2}\right)\)
\(1;x^2-x-2\)
\(=x^2-2x+x-2\)
\(=x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
\(2,x^2+6x-7\)
\(=x^2-x+7x-7\)
\(=x\left(x-1\right)+7\left(x-1\right)=\left(x+7\right)\left(x-1\right)\)