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23 tháng 7 2023

\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)

21 tháng 8 2023

a) \(x^8+x^4-2\)

\(=x^8+x^7+x^6+x^5+2x^4+2x^3+2x^2+2x-x^7-x^6-x^5-x^4-2x^3-2x^2-2x-2\)

\(=x\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)-\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)

\(=\left(x-1\right)\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)

\(=\left(x-1\right)\left[x^4\left(x^3+x^2+x+1\right)+2\left(x^3+x^2+x+1\right)\right]\)

\(=\left(x-1\right)\left(x^4+2\right)\left(x^3+x^2+x+1\right)\)

\(=\left(x-1\right)\left(x^2+2\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)\left(x+1\right)\)

21 tháng 8 2023

c) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=x^4+2x^3+x^2-2x^2-2x-15\)

\(=x^4+2x^3-x^2-2x-15\)

\(=x^4+x^3+3x^2+x^3+x^2+3x-5x^2-5x-15\)

\(=x^2\left(x^2+x+3\right)+x\left(x^2+x+3\right)-5\left(x^2+x+3\right)\)

\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)

`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:=2(x-2)+y(x-2)

=(x-2)(y+2)

b: \(=\left(x+y\right)^2-4\)

=(x+y+2)(x+y-2)

a: 6x-2y=2(3x-y)

b: =(x-y)(x-2)(x+2)

AH
Akai Haruma
Giáo viên
8 tháng 12 2021

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. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

23 tháng 5 2022

`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)`

23 tháng 5 2022

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) \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)

27 tháng 3 2021

Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!

27 tháng 3 2021

Trả lời:

A=(x-1)(x+2)(x-3)(x+4)-144

A= (x2-5x-14)(x2-5x-24)-144 (1)

đặt m=x2-5x-14

=> A= m.(m-10)-144

A=m2-10m-144

A= (m-18)(m+8)

thay m vào, ta có:

A= (x2-5x-32)(x2-5x-6)

A=(x2-5x-32)(x+1)(x-6)

28 tháng 9 2021

\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)

28 tháng 9 2021

\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)

19 tháng 8 2021

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: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+15\right)\left(3x-4\right)\)

13 tháng 11 2021

a) \(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)

b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)

c) đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+4\right)+2=t\left(t+3\right)+2=t^2+3t+2=\left(t^2+t\right)+\left(2t+2\right)=t\left(t+1\right)+2\left(t+1\right)=\left(t+1\right)\left(t+2\right)=\left(x^2+x+2\right)\left(x^2+x+3\right)\)