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1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
1.
a. \((x+1)(x^2-x+1)-(x-1)(x^2+x+1)\)
\(=x^3 + 1-(x^3-1) = 2 \)
b.
\(\dfrac{2x^2-4x+2}{2x-2}=\dfrac{2\left(x^2-2x+1\right)}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
2.
a. \(x^2-4y^2+12y-9=x^2-\left[\left(2y\right)^2-2\cdot2y\cdot3+3^2\right]=x^2-\left(2y-3\right)^2=\left(x-2y+3\right)\left(x+2y-3\right)\)
b.
\(5x^2+3\left(x+y\right)^2-5y^2\)
\(=3\left(x+y\right)^2+5\left(x^2-y^2\right)\)
\(=3\left(x+y\right)^2+5\left(x+y\right)\left(x-y\right)\)
\(=\left(x+y\right)\left[3\left(x+y\right)+5\left(x-y\right)\right]\)
\(=\left(x+y\right)\left(3x+3y+5x-5y\right)\)
\(=\left(x+y\right)\left(8x-2y\right)=2\left(x+y\right)\left(4x-y\right)\)