a) \(A=3x^2+x-1=3\left(x^2+\frac{x}{3}+\frac{1}{36}\right)-\frac{13}{12}=3\left(x+\frac{1}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\forall x\)

Dấu"=" xảy ra \(\Leftrightarrow x+\frac{1}{6}=0\)\(\Leftrightarrow x=-\frac{1}{6}\)

Vậy \(MinA=-\frac{13}{12}\Leftrightarrow x=-\frac{1}{6}\)

b)\(B=t^2-6t=\left(t^2-6t+9\right)-9=\left(t-3\right)^2-9\ge-9\forall t\)

Dấu "=" xảy ra \(\Leftrightarrow t-3=0\)\(\Leftrightarrow t=3\)

Vậy \(MinB=-9\Leftrightarrow t=3\)

c)\(C=x^2+\frac{3}{2}y^2-2x-4y+4\)

\(=\left(x^2-2x+1\right)+\frac{3}{2}\left(y^2-\frac{8}{3}y+\frac{16}{9}\right)+\frac{1}{3}\)

\(=\left(x-1\right)^2+\frac{3}{2}\left(y-\frac{4}{3}\right)^2+\frac{1}{3}\ge\frac{1}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-\frac{4}{3}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{4}{3}\end{cases}}\)

Vậy \(MinC=\frac{1}{3}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{4}{3}\end{cases}}\)

d)\(D=2x^2+y^2-2xy+4x+2024\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)+2020\)

\(=\left(x-y\right)^2+\left(x+2\right)^2+2020\ge2020\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\x+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y\\x=-2\end{cases}}\)\(\Leftrightarrow x=y=-2\)

Vậy \(MinD=2020\Leftrightarrow x=y=-2\)

ta có : 

1)

ta có: x+2y=1 => x=1-2y

thay vào bt, ta có:

\(A=\left(1-2y\right)^2+2y^2=1-4y+4y^2+2y^2=6y^2-4y+1\\ A=6\left(x-\dfrac{4}{2.6}\right)^2+\dfrac{4.6.1-\left(-4\right)^2}{4a}\ge\dfrac{4.6.1-\left(-4\right)^2}{46}=\dfrac{1}{3}\)

A đạt min khi x-1/3=0 => x=1/3

vậy MIN A=1/3 tại x=1/3

áp dụng bđt cô si cho 4 số ta có

\(x^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\ge4\sqrt[4]{x^4.\dfrac{1}{16}.\dfrac{1}{16}.\dfrac{1}{16}}\)

⇔ \(x^4+\dfrac{3}{16}\ge x.\dfrac{1}{2}\)

cmtt ta có

\(y^4+\dfrac{3}{16}\ge y\dfrac{1}{2}\)

cộng các vế của bđt trên ta có

\(x^4+y^4+\dfrac{3}{8}\ge\dfrac{1}{2}\left(x+y\right)\)

⇔ \(C+\dfrac{3}{8}\ge\dfrac{1}{2}\)

⇔ \(C\ge\dfrac{1}{8}\)

minC=\(\dfrac{1}{8}\) khi x=y=\(\dfrac{1}{2}\)