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c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
a) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)
b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)
a: \(=\dfrac{2}{5}\left(xy-x-y^2+1\right)\)
\(=\dfrac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]\)
\(=\dfrac{2}{5}\left(y-1\right)\left(x-y-1\right)\)
b: \(=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
1. ( x2 - x + 2 )4 - 3x2 ( x2 - x + 2 )2 + 2x4
Đặt t = x2 - x + 2 , ta có :
t4 - 3x2t2 + 2x4
= t4 - 2x2t2 - x2t2 + 2x4
= t2 ( t2 - 2x2 ) - x2 ( t2 - 2x2 )
= ( t2 - x2 ) ( t2 - 2x2 )
= ( t - x ) ( t + x ) ( t2 - 2x2 )
= ( x2 - x + 2 - x ) ( x2 - x + 2 + x ) [ ( x2 - x + 2 )2 - 2x2 ]
= ( x2 - 2x + 2 ) ( x2 + 2x ) ( x2 - 3x + 2 ) ( x2 + x + 2 )
2. 3 ( - x2 + 2x + 3 )4 - 26x2 ( - x2 + 2x + 3 )2 - 9x4
Đặt y = - x2 + 2x + 3 , ta có :
3y4 - 26x2y2 - 9x4
= x2y2 + 3y4 - 9x4 - 27x2y2
= y2 ( x2 + 3y2 ) - 9x2 ( x2 + 3y2 )
= ( y2 - 9x2 ) ( x2 + 3y2 )
= ( y - 3x ) ( y + 3x ) ( x2 + 3y2 )
= ( - x2 + 2x + 3 - 3x ) ( - x2 + 2x + 3 + 3x ) [ x2 + 3 ( - x2 + 2x + 3 )2 ]
= ( - x2 - x + 3 ) ( - x2 + 5x + 3 ) ( 3x4 - 12x3 - 5x2 + 36x + 27 )
\(a,=ab\left(a+3\right)\\ b,=\left(x-1\right)^2\\ c,=x\left[\left(x-3\right)^2-y^2\right]=x\left(x-y-3\right)\left(x+y-3\right)\)
Lời giải:
a. Không phân tích được nữa
b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$
$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$
`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`
`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`
`c, xy^2 + x^2y + 1/4y^3`
`= y(xy + x^2 + 1/4y^2)`
`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`
`= (x+1+y)(x+1-y)`
\(a,=2xy\left(2y-x\right)\\ b,=x^2\left(x-4\right)+5\left(x-4\right)=\left(x^2+5\right)\left(x-4\right)\\ c,=\left(x-y\right)\left(x^2-25\right)=\left(x-y\right)\left(x-5\right)\left(x+5\right)\)
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(t=x^2+5x+4\)
(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)
Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)
a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
