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\(4x^4-8x^3+3x^2-8x+4\)
\(=\left(4x^4-8x^3\right)+\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=4x^3\left(x-2\right)+3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(4x^3+3x-2\right)\)
\(8x^2y-18y\)
\(=2y\left(x^2-9\right)\)
\(=2y\left(x-3\right)\left(x+3\right)\)
\(3x^3+6x^2+3x-12xy^2\)
\(=3x\left(x^2+2x+1-4y^2\right)\)
\(=3x\left[\left(x+1\right)^2-\left(2y\right)^2\right]\)
\(=3x\left(x+1-2y\right)\left(x+1+2y\right)\)
Mình ko thêm bớt hạng tử nhé.
\(8x^3-3x+6x^2-1\)
\(=\left(8x^3-1\right)+\left(6x^2-3x\right)\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right)+3x\left(2x-1\right)\)
\(=\left(2x-1\right)\left[\left(4x^2+2x+1\right)+3x\right]\)
\(=\left(2x-1\right)\left(4x^2+5x+1\right)\)
\(=\left(2x-1\right)\left[4x\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(2x-1\right)\left(x+1\right)\left(4x+1\right)\)
\(8x^3-3x+6x^2-1=\left(8x^3-12x^2+6x-1\right)+\left(18x^2-9x\right)\)
\(=\left(\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\right)+\left(18x^2-9x\right)\)
\(=\left(2x-1\right)^3+9x\left(2x-1\right)=\left(2x-1\right)\left(\left(2x-1\right)^2+9x\right)\)
\(=\left(2x-1\right)\left(4x^2-4x+1+9x\right)=\left(2x-1\right)\left(4x^2+5x+1\right)\)
\(3x^4+6x^3-7x^2+8x-10\)
\(=\left(3x^4-3x^3\right)+\left(9x^3-9x^2\right)+\left(2x^2-2x\right)+\left(10x-10\right)\)
\(=\left(x-1\right)\left(3x^3+9x^2+2x+10\right)\)
1/\(9x^2+6x-575=\left(3x\right)^2+2.3x.1+1-576=\left(3x+1\right)^2-24^2=\left(3x-23\right)\left(3x+25\right)\)
2/\(81x^4+4=81x^4+36x^2+4-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)
3/đặt \(t=x^2+8x+7\) thì đa thức cần phân tích:
t(t+8)+15=t2+8t+15=t2+3t+5t+15=t(t+3)+5(t+3)=(t+3)(t+5)=(x2+8x+10)(x2+8x+12)=(x2+8x+10)(x2+2x+6x+12)
=(x2+8x+10)[x(x+2)+6(x+2)]=(x2+8x+10)(x+2)(x+6)
tạm thế này đã, phải đi ăn cơm rồi :v
a: =(2x-3y)^2-4(2x-3y)
=(2x-3y)(2x-3y-4)
b: =3x^2+21x-x-7
=(x+7)(3x-1)
c: =(3x-1)^4+2(3x-1)^2+1
=[(3x-1)^2+1]^2
d: =2x^3-2x^2-x^2+x+x-1
=(x-1)(2x^2-x+1)
Câu 1 :
\(a,x^3-6x^2+9x\)
\(=x\left(x^2-6x+9\right)\)
\(=x\left(x-3\right)\)
b;c tự lm nha !!! : câu 2 cx vậy
1.b) x2 - 2xy + 3x - 6y = x2 - 2xy + 3x - 3y x 2
= (x2 - 2xy) + (3x - 3y) x 2
= 2x (x - y) + 3 (x - y) x 2
= (x - y) (2x + 3 x 2)
= (x - y) (2x + 6)
2.
(2x4 - 3x3 + 3x2 - 3x + 1) : (x2 + 1)
2x4 - 3x3 + 3x2 - 3x + 1 / x2 + 1
2x4 + 2x2 / 2x2 - 3x + 1
0 - 3x3 + x2 - 3x + 1 /
- 3x3 - 3x /
0 + x2 + 0 + 1 /
x2 + 1 /
0
=> đây là phép chia hết
Vậy (2x4 - 3x3 + 3x2 - 3x + 1) : (x2 + 1) = 2x2 - 3x + 1
(Sai thì thôi)
\(a,3x-15xy=3x\left(1-5y\right)\\ ---\\ 8x^2+6x-4=2\left(4x^2+3x-2\right)\\ ---\\ 5x^2+25xy+10y^2=5\left(x^2+5xy+2y^2\right)\\ ---\\ 9x^2y^2+6x^2y-\dfrac{1}{2}xy^2=\dfrac{1}{2}xy\left(18xy+12x-y\right)\)
\(=8x^3-4x^2-2x^2+x+2x-1=\left(2x-1\right)\left(4x^2-x+1\right)\)