Phân tích đa thức thành nhân tử:
a, (x^2-x)^2+4(x^2-x)-12
b, (x+1)(x+2)(x+3)(x+4)-24
c,x^9-x^7+x^6-x^5-x^4+x^3-x^2+1
Phân tích đa thức thành nhân tử:
a. (x^2+x+1)(x^2+x+2)-12
b. (x^2+2x)^2-2(x^2+2x)-3
c. (x^2+x)^2+4x^2+4x-12
d. (x+2)(x+4)(x+6)(x+8)+16
a: =(x^2+x)^2+3(x^2+x)-10
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
b: (x^2+2x)^2-2(x^2+2x)-3
=(x^2+2x-3)(x^2+2x+1)
=(x+1)^2*(x+3)(x-1)
c: =(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
d: =(x^2+10x+16)(x^2+10x+24)+16
=(x^2+10x)^2+40(x^2+10x)+400
=(x^2+10x+20)^2
Phân tích các đa thức sau thành nhân tử:
a) \({\left( {x - 1} \right)^2} - 4\)
b) \(4{x^2} + 12x + 9\)
c) \({x^3} - 8{y^6}\)
d) \({x^5} - {x^3} - {x^2} + 1\)
e) \( - 4{x^3} + 4{x^2} + x - 1\)
f) \(8{x^3} + 12{x^2} + 6x + 1\)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
Phân tích đa thức thành nhân tử:
a) (x+1)(x+2)(x+3)(x+4)-15
b) ((2x+5)^2)-(x-9)^2
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: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
Bài 1:Phân tích đa thức thành nhân tử:
a,x^4+3x^3-6x^2-8x
b,x^3-y^3+z^3-3xyz
c,(x+1).(x+3).(x+5).(x+7)-5
`b)x^3+y^3+z^3-3xyz`
`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`
`=(x+y)^3+z^3-3xy(x+y+z)`
`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`
`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`
`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`
Bạn xem lại đề câu b) nhé đề đúng là `x^3+y^3+z^3-3xyz`.Còn được thì xem lại câu c) nhé do số quá xấu.
`a)x^4+3x^3-6x^2-8`
`=x(x^3+3x^2-6x-8)`
`=x[(x-2)(x^2+2x+4)+3x(x-2)]`
`=x(x-2)(x^2+5x+4)`
`=x(x-2)(x^2+x+4x+4)`
`=x(x-2)(x+1)(x+4)`
`c)(x+1)(x+3)(x+5)(x+7)-5`
`=[(x+1)(x+7)][(x+3)(x+5)]-5`
`=(x^2+8x+7)(x^2+8x+15)-5`
`=(x^2+8x+11)^2-4^2-5`
`=(x^2+8x+11)^2-21`
`=(x^2+8x+11-sqrt21)(x^2+8x+11+sqrt21)`
phân tích đa thức sau thành nhân tử:
a)x^4-y^4
b)x^2-3y^2
c)(3x-2y)^2-4(x+y)^2
d)9(x-y)^2-4(x+y)^2
f)x^3+27
g)27x^3-0,001
h)125x^3-1
\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)
phân tích đa thức thành nhân tử:a)x^5-x^3+x^2-1/5
b)5x^3-45x
c)16x^4y^2+2xy^5
d)a^3-8+6a^2-12a
e)x^4+x^3+x+1
a: Ta có: \(x^5-x^3+x^2-1\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x-1\right)\cdot\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
b: Ta có: \(5x^3-45x\)
\(=5x\left(x^2-9\right)\)
\(=5x\left(x-3\right)\left(x+3\right)\)
c: Ta có: \(16x^4y^2+2xy^5\)
\(=2xy^2\left(8x^3+y^3\right)\)
\(=2xy^2\cdot\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d: Ta có: \(a^3-8+6a^2-12a\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+8a+4\right)\)
e: Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
Phân tích đa thức thành nhân tử:
a) x - 2y + x^2- 4y^2
b) x^2 - 4x^2y^2 + y^2 + 2xy
c) x^6 - x^4 +2x^3 + 2x^2
d) x^3 + 3x^2 + 3x +1 - 8y^3
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(x+2y+1\right)\)
b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
c) Ta có: \(x^6-x^4+2x^3+2x^2\)
\(=x^4\left(x-1\right)\left(x+1\right)+2x^2\left(x+1\right)\)
\(=\left(x+1\right)\left[x^4\left(x-1\right)+2x^2\right]\)
\(=x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\)
\(=x^2\left(x+1\right)\cdot\left(x^3-x^2+2\right)\)
d) Ta có: \(x^3+3x^2+3x+1-8y^3\)
\(=\left(x+1\right)^3-\left(2y\right)^3\)
\(=\left(x+1-2y\right)\left[\left(x+1\right)^2+2y\left(x+1\right)+4y^2\right]\)
\(=\left(x-2y+1\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
Phân tích đa thức thành nhân tử:
a) x - 2y + x^2 - 4y^2
b) x^2 - 4x^2y^2 + y^2 + 2xy
c) x^6 - x^4 + 2x^3 + 2x^2
d) x^3 + 3x^2 + 3x + 1 - 8y^3
a, \(x-2y+x^2-4y^2=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(1+x+2y\right)\)
b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
Phân tích đa thức thành nhân tử:
a) 81x5-x3
b) 9x2y-12xy+4y
c) (5-x)2-16(x-2)2
d) 9x2-y2-21x-7y
e) -y2+8y+9x2-16
f) 5x2-4x-1
g) (x+7)(x+9)-17
h) x(x+2)(x+4)(x+6)+15
a: \(81x^5-x^3\)
\(=x^3\left(81x^2-1\right)\)
\(=x^3\left(9x-1\right)\left(9x+1\right)\)
b: \(9x^2y-12xy+4y\)
\(=y\left(9x^2-12x+4\right)\)
\(=y\left(3x-2\right)^2\)
c: \(\left(5-x\right)^2-16\left(x-2\right)^2\)
\(=\left(x-5\right)^2-\left(4x-8\right)^2\)
\(=\left(x-5-4x+8\right)\left(x-5+4x-8\right)\)
\(=-3\left(x-1\right)\left(5x-13\right)\)
d: Ta có: \(9x^2-y^2-21x-7y\)
\(=\left(3x-y\right)\left(3x+y\right)-7\left(3x+y\right)\)
\(=\left(3x+y\right)\left(3x-y-7\right)\)
e: Ta có: \(-y^2+8y-16+9x^2\)
\(=-\left(y^2-8y+16-9x^2\right)\)
\(=-\left(y-4-3x\right)\left(y-4+3x\right)\)
f: Ta có: \(5x^2-4x-1\)
\(=5x^2-5x+x-1\)
\(=5x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(5x+1\right)\)
phân tích đa thức thành nhân tử:A=(x-1)(x+2)(x-3)(x+4)-144
Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!
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)