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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)
1) \(x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
2)\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-x\right)\left(x+y+z\right)\)
3)\(x^3+y^3-3x-3y=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-3\right)\)
\(1.x^3+y^3-x-y=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
2.\(3\left(x^2+6xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
3.\(\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-3\right)\)
cho mình nha
a) \(y^2+2y^2-3y=y.y+y.2y-y.3\)
\(=y\left(y+2y-3\right)\)
b) \(2x^4-x^3+2x^2+1=2x^2.x^2-x^2.x+2x.x^2+1\)
\(=x^2\left(2x^2-x+2x\right)+1=x^2.x\left(2x-1+2\right)+1\)
k mình nha
a. y2+2y2-3y
=y(y+2y-3)
b. 2x4-x3+2x2+1
=2x2(x2+1)-(x3-1)
=2x2(x+1)(x-1)-(x-1)(x2+x+1)
=(x-1)[2x2(x+1)-(x2+x+1)]
=(x-1)(2x3+2x2-x2-x-1)
=(x-1)(2x3+x2-x-1)
1) \(\left(x^2+8x+7\right).\left(x+3\right).\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right).\left(x^2+5x+3x+15\right)+15\)
\(=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)
Ta đặt: \(x^2+8x+7=n\)
\(=n.\left(n+8\right)+15\)
\(=n^2+8n+15\)
\(=n^2+3n+5n+15\)
\(=\left(n^2+3n\right)+\left(5n+15\right)\)
\(=n.\left(n+3\right)+5.\left(n+3\right)\)
\(=\left(n+3\right).\left(n+5\right)\)
\(=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)
\(=\left(x^2+8x+10\right).[x.\left(x+2\right)+6.\left(x+2\right)]\)
\(=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)
2) \(x^2-2xy+3x-3y-10+y^2\)
\(=\left(x-y\right)^2+3.\left(x-y\right)-10\)
Ta đặt: \(x-y=n\)
\(=n^2+3n-10\)
\(=n^2-2n+5n-10\)
\(=\left(n^2-2n\right)+\left(5n-10\right)\)
\(=n.\left(n-2\right)+5.\left(n-2\right)\)
\(=\left(n-2\right).\left(n+5\right)\)
\(=\left(x-y-2\right).\left(x-y+5\right)\)
1)\(=x^2\left(x-y\right)-y\left(x-y\right)\\ =\left(x-y\right)\left(x^2-y\right)\)
2)\(=\left(x^2+x\right)-\left(2xy+2y\right)\\ =x\left(x+1\right)-2y\left(x+1\right)\\ =\left(x+1\right)\left(x-2y\right)\)
3)\(=\left(x^2+2.x.2y+4y^2\right)-y^2\\ =\left(x+2y\right)^2-y^2\\ =\left(x+2y+y\right)\left(x+2y-y\right)\)
1) x3 - x2y - xy + y2
= (x3 - x2y) - (xy - y2)
= x2.(x - y) - y.(x - y)
= (x - y).(x2 - y)
2) x2 - 2xy + x - 2y
= (x2 + x) - (2xy + 2y)
= x.(x + 1) - 2y.(x + 1)
= (x + 1).(x - 2y)
3) x2 + 4xy + 3y2
= x2 + 3xy + xy + 3y2
= (x2 + 3xy) + (xy + 3y2)
= x.(x + 3y) + y.(x + 3y)
= (x + 3y).(x + y)
1 ) \(x^3-x^2y-xy+y^2\)
\(=\left(x^3-x^2y\right)-\left(xy-y^2\right)\)
\(=x^2.\left(x-y\right)-y.\left(x-y\right)\)
\(=\left(x-y\right).\left(x^2-y\right)\)
2 ) \(x^2-2xy+x-2y\)
\(=\left(x^2+x\right)-\left(2xy+2y\right)\)
\(=x.\left(x+1\right)-2y.\left(x+1\right)\)
\(=\left(x+1\right).\left(x-2y\right)\)
3 ) \(x^2+4xy+3y^2\)
\(=x^2+3xy+xy+3y^2\)
\(=\left(x^2+3xy\right)+\left(xy+3y^2\right)\)
\(=x.\left(x+3y\right)+y.\left(x+3y\right)\)
\(=\left(x+3y\right).\left(x+y\right)\)
3x^2 +3y^2 -6xy -12
=3(x^2 - 2xy +y^2 - 2^2 )
=3 (x-y)^2 - 2^2
=3(x-y-2)(x-y+2)
3(x+y) -(x^2+2xy+y^2)
=3(x+y) -(x+y)^2
(x+y)(3-x-y)
\(x^3+y^3+3y^2+3y+1\\ =x^3+\left(y+1\right)^3\\ =\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\\ =\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\\ =\left(x+y+1\right)\left(x^2+y^2+2y+1-xy-x\right)\)
=(x^3+3y+3y^2+y^3)+1
=(x+y)^3+1
=(x+y+1).[(x+y)^2+(x+y)+1]