C

C

Thay x=-1 vào pt ta có:

\(3m\left(-1\right)^2-2.\left(-1\right)=4\\ \Leftrightarrow3m+2-4=0\\ \Leftrightarrow3m-2=0\\ \Leftrightarrow m=\dfrac{2}{3}\)

Chọn A

B

Diện tích mặt bàn là:
\(\left(14:2\right)\times\left(14:2\right)\times3,14=153,86\left(dm^2\right)\)

Chọn C

Bài 1:
\(a,\left(2x+5\right)\left(y-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+5=0\\y-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\y=1\end{matrix}\right.\\ n,x^2+3x=0\\ \Leftrightarrow x\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Bài 2:
a, Ta có:\(A=\left(x-1\right)^2+\left(y+3\right)^2+2021\ge2021\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Vậy \(A_{min}=2021\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

b, Ta có: \(B=\left(x+2\right)^2+\left(y-\dfrac{1}{5}\right)^2-10\ge-10\)