Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)
c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=x^4-14x^2-32\)
\(=\left(x^2-16\right)\left(x^2+2\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)
\(=\left(x^2+5x+1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`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)`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(9x^2-\dfrac{1}{4}\)
\(=\left(3x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
d) \(\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
e) \(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3+x-y\right)\left(3-x+y\right)\)
f) \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$
$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, 9x^2 - 16 = (3x+4)(3x-4)`
`b, 4x^2 - 12xy + 9y^2 = (2x-3y)^2`
`c, t^3-8 = (t-2)(t^2 - 2t + 4)`
`d, 2ax^3y^3 + 2a = 2a(x^3y^3 + 1) = 2a(xy+1)(x^2y^2 - xy + 1)`
a) \(\left(9x^2-16\right)=\left(3x-4\right)\left(3x+4\right)\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(t^3-8=\left(t-2\right)\left(t^2+2t+4\right)\)
d) \(2ax^3y^3+2a=2a\left(x^3y^3+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)(x+2)^2+2(x^2-4)+(x-2)^2`
`=(x+2)^2+2(x-2)(x+2)+(x-2)^2`
`=(x+2+x-2)^2=(2x)^2=4x^2`
`b)x^2-x+1/4`
`=x^2-2.x .1/2+1/4=(x-1/2)^2`
`c)(x+y)^3-(x-y)^3`
`=(x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]`
`=2y(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)`
`=2y(3x^2+y^2)`
a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)
\(=\left(x+2\right)^2+2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(x+2+x-2\right)^2=\left(2x\right)^2=4x^2\)
b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
c) \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
a) (x-2)(x+2)(x^2-10)-72=(x^2-4)(x^2-82)
b) x^8+x^6+x^4+x^2+1=x^2 (x^4+x^3+x^2+1+1/x^2)
c)(x+y)^4+x^4+y^4=(x+y)^4+(x+y)^4=2 (x+y)^4
a) (x-2)(x+2)(x^2 - 10) -72
= (x^2 - 4)(x^2 - 10) - 72
= x^4 - 4x^2 -10x^2 + 40 - 72
= x^4 - 14x^2 - 32
= x^4 - 16x^2 + 2x^2 - 32
= x^2(x^2 - 16) + 2(x^2 - 16)
= (x^2 - 16)(x^2 + 2)
= (x-4)(x+4)(x^2 + 2)
c) (x+y)4 + x4 + y4
= 2x4 + 4xy3 + 6x2y2 + 4x3y + 2y3
= 2(y4 + 2xy3 + 3x2y2 + 2x3y + x4)
= 2(y2 + xy + y2)2