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a/ \(=3y^2-6y-2x+1\)
b/ \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
c/ \(=\left(2-x\right)^3\)
d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)
\(=xy\left(y+x+3\right)+x^2\left(y+x+3\right)-3x\left(y+x+3\right)\)
\(=\left(xy+x^2-3x\right)\left(y+x+3\right)=x\left(y+x-3\right)\left(y+x+3\right)\)
e/ \(=xy-x^2+2x-y^2+xy-2y\)
\(=x\left(y-x+2\right)-y\left(y-x+2\right)=\left(x-y\right)\left(y-x+2\right)\)
a) =(2x+3y-1)2
b)=-(x-1)3
c)=-(x3-6x2+12x-8)=-(x-2)3
d)x3 + 2x2y + xy2 – 9x
= x(x2 + 2xy + y2 -9)
= x[(x2 + 2xy + y2) - 32]
= x[(x + y)2 - 32]
= x (x + y – 3)(x + y + 3)
e) 2x-2y-x2+2xy-y2=2(x-y)-(x-y)2=(x-y)(2-x+y)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a, \(x^3-6x^2+9x\)
\(=x\left(x^2-6x+9\right)\)
\(=x \left(x-3\right)\)
Câu b, c cũng tượng tự nha bn , dễ mà
#hoc_tot#
b) \(x^2-2xy+3x-6y=x\left(x-2y\right)+3\left(x-2y\right)=\left(x-2y\right)\left(x+3\right)\)
c)\(x^2-8x+7=x^2-x-7x+7=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\)
a)\(x^3-6x^2+9x=x\left(x^2-2\cdot x\cdot3+3^2\right)=x\left(x-3\right)^2\)
~ Chúc bạn học tốt ~
4x2 -6x= 2x(2x-3)
b) 3x3 -6x2y -24xy + 12x2 = \(3x\left(x^2-2xy-8y+4x\right)\)
c) x2 -25 + y2 + 2xy\(=x^2+2xy+y^2-25\)\(=\left(x+y\right)^2-5^2\)
=>\(\left(x+y+5\right)\left(x+y-5\right)\)
a) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)
b) \(=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)
f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=x^2+2xy+y^2-x^2+y^2\)
\(=2y^2+2xy\)
\(=2y\left(x+y\right)\)
c) \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-x^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)
\(=\left(4x^2-1\right)\left(y^2-1\right)\)
\( a){x^3} + 3{x^2} + 3x + 1\\ = {x^3} + 3.{x^2}.1 + 3.x{.1^2} + {1^3}\\ = {\left( {x + 1} \right)^3}\\ b){x^3} - 6{x^2} + 12x - 8\\ = {x^3} - 3.{x^2}.2 + 3.x{.2^2} - {2^3}\\ = {\left( {x - 2} \right)^3}\\ c){x^2} - 2xy + {y^2} - 16\\ = {\left( {x - y} \right)^2} - 16\\ = \left( {x - y - 4} \right)\left( {x - y + 4} \right)\\ d)49 - {x^2} + 2xy - {y^2}\\ = 49 - \left( {{x^2} - 2xy + {y^2}} \right)\\ = 49 - {\left( {x - y} \right)^2}\\ = \left[ {7 - \left( {x - y} \right)} \right]\left[ {7 + \left( {x - y} \right)} \right]\\ = \left( {7 - x + y} \right)\left( {7 + x - y} \right) \)
\(x^3+3x^2+3x+1=\left(x^3+1\right)+3\left(x^2+x\right)=\left(x+1\right)\left(x^2-x+1\right)+3x\left(x+1\right)=\left(x+1\right)\left(x^2-x+3x+1\right)=\left(x+1\right)\left(x^2+2x+1\right)=\left(x+1\right)\left(x+1\right)^2=\left(x+1\right)^3\)
\(x^3-6x^2+12x-8=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)=x^2\left(x-2\right)-4x\left(x-2\right)+4\left(x-2\right)=\left(x-2\right)\left(x^2-4x+4\right)=\left(x-2\right)\left(x-2\right)^2=\left(x-2\right)^3\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
\(49-x^2+2xy-y^2=49-\left(x-y\right)^2=\left(7-x+y\right)\left(7+x-y\right)\)
\(x^2+7x+12=x^2+3x+4x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+4\right)\left(x+3\right)\)