biểu thức <=> x³-3x²+9x+3x²-9x+27-54=0

<=> x³- 27 =0

<=> x=3

\(\left(x+3\right)\cdot\left(x^2-3x+9\right)-54=0\)

\(x^3+27=54\)

\(x^3=54-27=27\)

\(\Rightarrow x=3\)

(x+3)(x²-3x+3²)-54=0

x³+3³-54=0

x³-27=0

x³=27=3³

=>x=3

Vậy x=3

dmm

a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)

\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)

\(\Leftrightarrow-11x^2+23x=0\)

\(\Leftrightarrow x\left(-11x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)

nhanh giùm mình cái

3 nhé bạn

a, \(\left(2x-1\right)\left(x+3\right)-2x^2+5x=7\)

\(\Leftrightarrow2x^2+6x-x-3-2x^2+5x=7\)

\(\Leftrightarrow2x^2+5x-3-2x^2+5x=7\)

\(\Leftrightarrow10x-10=0\Leftrightarrow x=1\)

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-4\right)\left(x+4\right)=54\)

\(\Leftrightarrow\left(x^3+27\right)-x\left(x^2-16\right)=54\)

\(\Leftrightarrow x^3+27-x^3+16x=54\)

\(\Leftrightarrow-27+16x=0\Leftrightarrow x=\frac{27}{16}\)