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c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
Lời giải:
Áp dụng HĐT: $(a-b)^3=a^3-b^3-3ab(a-b)$ cho cả hai bạn.
a.
$M=x^3-1-3x(x-1)-3x(x-1)^2+3x^2(x-1)+x^3$
$=2x^3-1+3x(x-1)[-1-(x-1)+x]$
$=2x^3-1+3x(x-1).0=2x^3-1$
b.
$D=[(x-y)-x]^3=-y^3$
\(M=3\left(3x+1\right)\left(9x^2-3x+1\right)-\left(x^3+1\right)\)
\(=3\left(27x^3+1\right)-x^3-1=80x^3+2=80.\left(\dfrac{1}{2}\right)^3+2=12\)
Sửa đề: \(N=\left(3x+1\right)\left(9x^2-3x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(N=27x^3+1-x^3-1=26x^3=26.10^3=26000\)
\(6,\Rightarrow6x^2+12x+6-2x^3-6x^2-6x-2+2x^3-2=1\\ \Rightarrow6x=-1\Rightarrow x=-\dfrac{1}{6}\\ 7,\Rightarrow\left(x-2\right)\left(x-2-x-2\right)=0\\ \Rightarrow-4\left(x-2\right)=0\\ \Rightarrow x-2=0\Rightarrow x=2\\ 8,\Rightarrow\left(x-2\right)^2-25=0\\ \Rightarrow\left(x-2-5\right)\left(x-2+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
c)
\(x^3-3.x^2.6+3.x.6^2-6^3=0\)
\(\left(x-6\right)^3=0\)
x-6=0
x=6
d)
\(x^3-3.x^2.1+3.x.1^2-1-x^3-3x-2=0\)
\(x^3-3x^2+3x-1-x^3-3x^2-2=0\)
\(-6x^2-3=0\)
\(-3\left(2x^2+1\right)=0\)
\(2x^2+1=0\)
2x2=-1
x2=1/2
x=\(\dfrac{\sqrt{2}}{2}\)
\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)