x = 3

a) x = 3.     

b) x Î {-3; 3}. 

c) x = 5.

d) x Î {-7; 7). 

a: \(21x^2-7x\left(3x-2\right)=42\)

\(\Leftrightarrow14x=42\)

hay x=3

\(\left(x+3\right)\left(x^2-3x+9\right)=7x^3+21x\\ \Leftrightarrow x^3+27=7x^3+21x\\ \Leftrightarrow6x^3+21x-27=0\\ \Leftrightarrow\left(6x^3-6x^2\right)+\left(6x^2-6x\right)+\left(27x-27\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x^2+6x+27\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x^2+6x+27=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\6\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{51}{2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\6\left(x+\dfrac{1}{2}\right)^2+\dfrac{51}{2}=0\left(vô.lí\right)\end{matrix}\right.\)

Vậy \(x=1\)

\(\Leftrightarrow x^3+27-7x^3-21x=0\)

\(\Leftrightarrow-6x^3-21x+27=0\)

\(\Leftrightarrow-6x^3+6x-27x+27=0\)

\(\Leftrightarrow-6x\left(x-1\right)\left(x+1\right)-27\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x^2+6x+27\right)=0\)

hay x=1

\(x=\sqrt{\sqrt{3}-\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

\(=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}\)

=1

Thay x=1 vào P, ta được:
\(P=\left(21\cdot1^2+6\cdot1\right)^{2017}=27^{2017}\)

45Nguyễn Đức Mạnh

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(VP=\frac{a\left(x^2-x-3\right)+b\left(x^2-2x-3\right)+c\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x-3\right)}\)

\(=\frac{\left(a+b+c\right)x^2+x\left(-a-2b+3c\right)+\left(-3a-3b+2c\right)}{\left(x+1\right)\left(x+2\right)\left(x-3\right)}\)

đồng nhất hệ số ta có 

\(\hept{\begin{cases}a+b+c=21\\-a-2b+3c=4\\-3a-3b+2c=-41\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=24\\b=\frac{-37}{5}\\c=\frac{22}{5}\end{cases}}\)