Giúp mình câu này nha x3+3x2+3x+1
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Lời giải:
1.
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$
$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$
2.
$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$
3.
$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$
$=(x-1)(x^2+3x+1)$
\(\Leftrightarrow\left(x+1\right)^3=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
a)=\(3x^3-15x^2+21x\)
b)\(=-2x^4y-10x^2y+2xy\)
c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)
d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)
e)\(=x^2-4y^2\)
f)\(=-2x^2y^3+y-3\)
g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)
h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)
i)\(=x^2-x-3\)
j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
\(x^3+3x^2+3x=-\dfrac{7}{8}\\ x^3+3x^2+3x+1=1-\dfrac{7}{8}\\ \left(x+1\right)^3=\dfrac{1}{8}\\ x+1=\dfrac{1}{2}\\ x=-\dfrac{1}{2}\)
Ta có: \(x^3+3x^2+3x=\dfrac{-7}{8}\)
\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+1=\dfrac{1}{2}\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
\(x^3+3x^2+3x+1=\left(x+1\right)^3\)
Cảm ơi bạn