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1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
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) \(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
b) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+\left(3x^2+3xy^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+3xy-1\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
Bài làm ai trên 11 điểm tích mình thì mình tích lại
Ông tùng hơn tùng số tuổi là :
29 + 32 = 61 (tuổi )
Vậy ông của tùng hơn tùng 61 tuổi
b) \(B=\left(2x^2+x-2\right)\left(2x^2+x-3\right)-12\)
Đặt \(2x^2+x-2=t\)
Ta được:
\(B=t\left(t-1\right)-12\)
\(B=t^2-t-12\)
\(B=t^2+3t-4t-12\)
\(B=t\left(t+3\right)-4\left(t+3\right)\)
\(B=\left(t+3\right)\left(t-4\right)\)
\(B=\left(2x^2+x+1\right)\left(2x^2+x-6\right)\)
\(B=\left(2x^2+x+1\right)\left(2x^2+4x-3x-6\right)\)
\(B=\left(2x^2+x+1\right)\left[2x\left(x+2\right)-3\left(x+2\right)\right]\)
\(B=\left(2x^2+x+1\right)\left(x+2\right)\left(2x-3\right)\)
Vậy...
`#040911`
`a)`
`x^2 + y^2 + 2xy - 25`
`= (x^2 + 2xy + y^2) - 25`
`= [ (x)^2 + 2*x*y + (y)^2] - 5^2`
`= (x + y)^2 - 5^2`
`= (x + y - 5)(x + y + 5)`
`b)`
`x^2 + 2x - 15`
`= x^2 + 5x - 3x - 15`
`= (x^2 + 5x) - (3x + 15)`
`= x(x + 5) - 3(x + 5)`
`= (x - 3)(x + 5)`
`c)`
`x^2 - x - 2`
`= x^2 - 2x + x - 2`
`= (x^2 - 2x) + (x - 2)`
`= x(x - 2) + (x - 2)`
`= (x + 1)(x - 2)`
`d)`
`3x^2 - 11x + 6`
`= 3x^2 - 9x - 2x + 6`
`= (3x^2 - 9x) - (2x - 6)`
`= 3x(x - 3) - 2(x - 3)`
`= (3x - 2)(x - 3)`
`a, (x+y)^2-25 = (x+y+5)(x+y-5)`.
`b, x^2+2x-15 = (x+1)^2-16 = (x-3)(x+5)`.
`c, x^2-x-2=(x-2)(x+1)`
`d, 3x^2-11x+6 = (3x-2)(x-3)`.