phân tích đa thức thành nhân tử:
a. 30a3 - 18a2b - 72b + 120a
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b. \(a^3+6a^2+12a+8\\ =a^3+3.a^2.2+3.2^2.a+2^3\\ =\left(a+2\right)^3\)
a: \(=30a\left(a^2+4\right)-18b\left(a^2+4\right)\)
\(=6\left(a^2+4\right)\left(5a-3b\right)\)
b: \(=\left(a+2\right)^3\)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)
\(=\left(x^2+5x+1\right)^2\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`
7a/
$x^3y+x-y-1=(x^3y-y)+(x-1)=y(x^3-1)+(x-1)$
$=y(x-1)(x^2+x+1)+(x-1)=(x-1)[y(x^2+x+1)+1]$
$=(x-1)(x^2y+xy+y+1)$
7b/
$x^2(x-2)+4(2-x)=x^2(x-2)-4(x-2)=(x-2)(x^2-4)$
$=(x-2)(x-2)(x+2)=(x-2)^2(x+2)$
7c/
$x^3-x^2-20x=x(x^2-x-20)=x[(x^2+4x)-(5x+20)]$
$x[x(x+4)-5(x+4)]=x(x+4)(x-5)$
7d/
$(x^2+1)^2-(x+1)^2=[(x^2+1)-(x+1)][(x^2+1)+(x+1)]$
$=(x^2-x)(x^2+x+2)$
$=x(x-1)(x^2+x+2)$
7e/
$6x^2-7x+2=(6x^2-3x)-(4x-2)=3x(2x-1)-2(2x-1)=(2x-1)(3x-2)$
7f/
$x^4+8x^2+12=(x^4+6x^2)+(2x^2+12)=x^2(x^2+6)+2(x^2+6)$
$=(x^2+6)(x^2+2)$
7g/
$(x^3+x+1)(x^3+x)-2=(t+1)t-2$ (đặt $x^3+x=t$)
$=t^2+t-2=(t^2+2t)-(t+2)=t(t+2)-(t+2)$
$=(t+2)(t-1)=(x^3+x+2)(x^3+x-1)$
$=[(x^3+x^2)-(x^2+x)+(2x+2)](x^3+x-1)$
$=[x^2(x+1)-x(x+1)+2(x+1)](x^3+x-1)$
$=(x+1)(x^2-x+2)(x^3+x-1)$
=( 30a3+120a)-(18a2b+72b)
=30a(a2+1)-18b(a2+1)
=(30a-18b)(a2+1)
=6(5a-3b)(a2+1)
120 a để 30a làm thừa số chung thì phải còn 4 chứ nhỉ ???