Thay \(x=-\dfrac{2}{3}\) vào A ta có:

\(A=6\cdot\left(-\dfrac{2}{3}\right)^3-3\cdot\left(-\dfrac{2}{3}\right)^2+2\cdot\left|-\dfrac{2}{3}\right|+4\)

\(A=6\cdot-\dfrac{8}{27}-3\cdot\dfrac{4}{9}+2\cdot\dfrac{2}{3}+4\)

\(A=-\dfrac{16}{9}-\dfrac{4}{3}+\dfrac{4}{3}+4\)

\(A=\dfrac{-16}{9}+4\)

\(A=\dfrac{20}{9}\)

Vậy: ....

A = 20/9

a. 6x³-x²-486x+81

= 6x³-54x²+53x²-477x-9x+81

= 6x².(x-9)+53x.(x-9)-9.(x-9)

= (x-9).(6x²+53x-9)

= (x-9)(6x²+54x-x-9)

=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)

b. x³-5x²+3x+9

= x³+x²-6x²-6x+9x+9

=x².(x+1)-6x.(x+1)+9.(x+1)

=(x+1)(x²-6x+9)=(x+1)(x-3)²

c. x³+3x²+6x+4

= x³+x²+2x²+2x+4x+4

= x².(x+1)+2x.(x+1)+4.(x+1)

= (x+1)(x²+2x+4)

d. 

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

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

\(=x^4-16-x^4+9=-7\)

b: \(=27x^3-8-27x^3+6=-2\)

c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)

1/ \(x^4+x^2-2=0\)

\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

2/ \(x^3+3x^2+6x+4=0\)

\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)

\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))

\(\Leftrightarrow x=-1\).

3/ \(x^3-6x^2+8x=0\)

\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)

4/ \(x^4-8x^3-9x^2=0\)

\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)

3: \(\Leftrightarrow a-15=0\)

hay a=15