Phân tích đa thức sau thành nhân tử:
a. A1 = (x + 2)4 + (x + 4)4 - 82
b. B1 = (x + 12)(x + 2)(x + 6)(x + 4) - 165x2
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\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
a) \(x^8+x^4-2\)
\(=x^8+x^7+x^6+x^5+2x^4+2x^3+2x^2+2x-x^7-x^6-x^5-x^4-2x^3-2x^2-2x-2\)
\(=x\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)-\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)
\(=\left(x-1\right)\left(x^7+x^6+x^5+x^4+2x^3+2x^2+2x+2\right)\)
\(=\left(x-1\right)\left[x^4\left(x^3+x^2+x+1\right)+2\left(x^3+x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^4+2\right)\left(x^3+x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+2\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)\left(x+1\right)\)
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)
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. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$
$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$
`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) \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
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)
\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)
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: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
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 \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+4\right)+2=t\left(t+3\right)+2=t^2+3t+2=\left(t^2+t\right)+\left(2t+2\right)=t\left(t+1\right)+2\left(t+1\right)=\left(t+1\right)\left(t+2\right)=\left(x^2+x+2\right)\left(x^2+x+3\right)\)