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Bài 1:
a. $x(x^2-5)=x^3-5x$
b. $3xy(x^2-2x^2y+3)=3x^3y-6x^3y^2+9xy$
c. $(2x-6)(3x+6)=6x^2+12x-18x-36=6x^2-6x-36$
d.
$(x+3y)(x^2-xy)=x^3-x^2y+3x^2y-3xy^2=x^3+2x^2y-3xy^2$
Bài 2:
a.
\((2x+5)(2x-5)=(2x)^2-5^2=4x^2-25\)
b.
\((x-3)^2=x^2-6x+9\)
c.
\((4+3x)^2=9x^2+24x+16\)
d.
\((x-2y)^3=x^3-6x^2y+12xy^2-8y^3\)
e.
\((5x+3y)^3=(5x)^3+3.(5x)^2.3y+3.5x(3y)^2+(3y)^3\)
\(=125x^3+225x^2y+135xy^2+27y^3\)
f.
\((5-x)(25+5x+x^2)=5^3-x^3=125-x^3\)
a) \(\Rightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)
\(\Rightarrow x^3-9x=0\)
\(\Rightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x^3-1=x^3-9x^2+2x^2+6\)
\(\Rightarrow7x^2=7\)
\(\Rightarrow x^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(1,\\ a,=x^3-5x\\ b,=3x^3y-6x^3y^2+9xy\\ c,=6x^2-6x-36\\ d,=x^3+2x^2y-3xy^2\\ 2,\\ a,=4x^2-25\\ b,=x^2-6x+9\\ c,=9x^2+24x+16\\ d,=x^3-6x^2y+12xy^2-8y^3\\ e,=125x^3+225x^2y+135xy^2+27y^3\\ f,=125-x^3\)
\(g,=8y^3+x^3\\ 3,\\ a,=x\left(x+2\right)\\ b,=\left(x-3\right)^2\\ c,=\left(x-y\right)\left(y+5\right)\\ d,=2x\left(y+1\right)-y\left(y+1\right)=\left(2x-y\right)\left(y+1\right)\\ e,=6x^2y^2\left(xy^2+2y-3x\right)\)
a: \(\left(x^2+1\right)\left(x-3\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+1-x^2-3x-9\right)\)
=(x-3)(-3x-8)
b: \(\left(x+2\right)^2+x\left(x+5\right)\)
\(=x^2+4x+4+x^2+5x\)
\(=2x^2+9x+4=2x^2+8x+x+4\)
=2x(x+4)+(x+4)
=(x+4)(2x+1)
c: \(\left(5x+4y\right)\left(5x-4y\right)-24x^2+15y^2\)
\(=25x^2-16y^2-24x^2+15y^2\)
\(=x^2-y^2=\left(x-y\right)\left(x+y\right)\)

\(\dfrac{x+1}{x}=3\Leftrightarrow x+1=3x\Leftrightarrow x=\dfrac{1}{2}\\ \dfrac{x^3+1}{x^3}=\dfrac{\dfrac{1}{8}+1}{\dfrac{1}{8}}=\dfrac{9}{8}\cdot8=9\)
cảm ơn bạn nhưng nó là x+ 1/x í ạ
\(x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=3^3-3\cdot3=27-9=18\)