A /B/C/D

Lời giải:
$6x^2-x-2=0$

$\Leftrightarrow x^2-\frac{x}{6}-\frac{1}{3}=0$

$\Leftrightarrow (x^2-\frac{x}{6}+\frac{1}{12^2})-\frac{49}{144}=0$

$\Leftrightarrow (x-\frac{1}{6})^2=\frac{49}{144}$

$\Rightarrow x-\frac{1}{6}=\frac{7}{12}$ hoặc $x-\frac{1}{6}=\frac{-7}{12}$

$\Rightarrow x=\frac{3}{4}$ hoặc $x=\frac{-5}{12}$

a)

\(A=\left(2x\right)^3-1-7x^3-7=8x^3-1-7x^3-7=x^3-8\)

b)

Thay \(x=-\dfrac{1}{2}\) vào biểu thức A:

\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{65}{8}\)

Ta có:

\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)

nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)

Thay \(x=4;y=1\) vào \(P\), ta được:

\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)

\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)

\(=1-1=0\)

Vậy \(P=0\) khi \(x=4;y=1\).