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12x^3+16x^2-5x-3 = 12x3+18x2 -2x2-3x -2x -3
= 12x2(x +3/2) -2x(x+3/2) -2(x+3/2)
=(12x2-2x-2)(x+3/2)
=2(6x2-x-1)(x+3/2)
=2[6x(x+1/3)-3(x+1/3)](x+3/4)
=2(6x-3)(x+1/3)(x+3/4)
=6(2x-1)(x+1/3)(x+3/4)
16 - 5x2 - 3
= -5x2 + 15x + x - 3
= -5x(x - 3) + (x - 3)
= (1 - 5x)(x - 3)
16x-5x^2-3=-5x^2+15x+x-3=5x(3-x)-(3-x) =(3-x)(5x-1) (xy+1)^2-x^2-2xy-y^2=(xy+1)^2-(x^2+2xy+y^2) =(xy+1)^2-(x+y)^2 =(xy+1-x-y)(xy+1+x+y) =[(xy-x)-(y-1)][(xy+x)+(y+1)] =[x(y-1)-(y-1)][x(y+1)+(y+1)] =(y-1)(x-1)(y+1)(x+1)
16x - 5x² - 3 = -5x² + 16x - 3 = -5x² + 15x + x - 3
= -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x)
\(-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(1-5x\right)\)
16x - 5x^2 - 3
=x - 5x^2 + 15x - 3
=x(1-5x) + 3(5x-1)
=3(5x-1) -x(5x-1)
=(3-x)(5x-1)
= (12x^3-6x^2)+(22x^2-11x)+(6x-3)
= (2x-1).(6x^2+11x+3)
= (2x-1).[(6x^2+2x)+(9x+3)]
= (2x-1).(3x+1).(2x+3)
k mk nha
Ta có: \(12x^3+16x^2-5x-3\)
\(=12x^3-6x^2+22x^2-11x+6x-3\)
\(=6x^2\left(2x-1\right)+11x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(6x^2+11x+3\right)\)
\(=\left(2x-1\right)\left(6x^2+9x+2x+3\right)\)
\(=\left(2x-1\right)\left(2x+3\right)\left(3x+1\right)\)
áp dụng hằng đẳng thức 7 và 3 nha dễ dàng mà . bn tách hết cho về mũ 3 hoặc mũ 2 nha
1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3xy\left(x-y\right)-12\left(x-y\right)\)
\(=\left(3xy-12\right)\left(x-y\right)\)
2) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
Ta có : 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 4.3.(x - y)
= 3(x - y)(x + y - 4)
\(16x-5x^2-3=\left(-5x^2+15x\right)+\left(x-3\right)=-5x.\left(x-3\right)+\left(x-3\right)\\ \)
\(=\left(x-3\right).\left(1-5x\right)\)
\(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x+2\right)\)
\(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-\left(5x^2-16x+3\right)\)
\(=-\left(5x^2-15x-x+3\right)\)
\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)
\(=-\left[\left(5x-1\right)\left(x-3\right)\right]\)
Câu 2:
\(x^2-x-6\)
\(=x^2+2x-3x-6\)
\(=x\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x-3\right)\left(x+2\right)\)
\(12x^3+16x^2-5x-3=12x^3+12x^2-9x+4x^2+4x-3\)
\(=3x\left(4x^2+4x-3\right)+\left(4x^2+4x-3\right)\)
\(=\left(3x+1\right)\left(4x^2+4x-3\right)=\left(3x+1\right)\left(4x^2+6x-2x-3\right)\)
\(=\left(3x+1\right)\left(2x\left(2x+3\right)-\left(2x+3\right)\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left(2x+3\right)\)
Mơn bn nà❤