\(8x^3-36x^2+54x-27\)

\(=\left(2x\right)^3-3.\left(2x\right)^2.3+3.2x.3^2-3^3\)

\(=\left(2x-3\right)^3\)

\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)

\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)

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

a, \(\dfrac{27}{8x^3-1}:\dfrac{3}{2x-1}\)

\(=\dfrac{27}{\left(2x-1\right)\left(4x^2+2x+1\right)}.\dfrac{2x-1}{3}\)

\(=\dfrac{9}{4x^2+2x+1}\)

b, \(\dfrac{8x^3+36x^2+54x+27}{2x+3}=\dfrac{\left(2x+3\right)^3}{2x+3}=\left(2x+3\right)^2\)

a) Sửa đề: \(8x^3+36x^2+54x+27\)

Ta có: \(8x^3+36x^2+54x+27\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=\left(2x+3\right)^3\)

b) Ta có: \(x^2+4x+4\)

\(=x^2+2\cdot x\cdot2+2^2\)

\(=\left(x+2\right)^2\)

Chúc bạn học tốt

cảm ơn bạn nha Nguyễn Như Quỳnh

Tại x = 103/2 ta có :

\(M=5.\left(\dfrac{103}{2}\right)^3-36.\left(\dfrac{103}{2}\right)^2+54.\dfrac{103}{2}+27=590281,375\)

8x³+36x²+54x+27

tại x =-4

=>8×(-4)³+36×(-4)²+54×(-4)+27

=8×(-64)+36×16+54×(-4)+27

=-512+576-216+27

=-125

(4x-3)(16x²+12x+9)-x²(64x-4)

=4x(16x²+12x+9)- 3(16x²+12x+9)-x²(64x-4)

=(64x³+48x²+36x)-(48x²+36x+27)-(64x³-4x²)

=64x³+48x²+36x-48x²-36x-27-64x³+4x²

=(64x³-64x³)+(48x²-48x²+4x²)+(36x-36x)-27

=4x²-27

tại x=-1/4

=> 4×(-1/4)²-27

=4×1/16-27

=1/4-27

=-107/4

(ko bt cs đúng ko nx )

Bài 1: 

a: \(C=\left(x-3\right)\left(x+3\right)-\left(x+5\right)\left(x-1\right)\)

\(=x^2-9-\left(x^2+4x-5\right)\)

\(=x^2-9-x^2-4x+5=-4x-4\)

b: \(D=\left(3x-2\right)^2+2\left(x+1\right)\left(3x-2\right)+\left(x+1\right)^2\)

\(=\left(3x-2+x+1\right)^2=\left(4x-1\right)^2=16x^2-8x+1\)

1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(2x-1\right)^3=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

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

\(\Leftrightarrow\left(x-2\right)^3=3^3\)

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)