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=\(\dfrac{1}{2}\)

Ai kết bạn với mink mink k cho hứa đó please

mình kết bn rùi đó k cho mình đi hihi

A = x²+2y² 

    = (x²+2x+1) + 2(y²+2y+1) -2x-4y-3

   = (x+1)² + 2(y+1)² - 2(x+2y) -3

   = (x+1)² + 2(y+1)² -5

 _{\(\Rightarrow A_{min}=\)}\(-5\Leftrightarrow\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\) 

a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)

\(xy\ge2\left(y-1\right)\ge0\Rightarrow x\ge\dfrac{2\left(y-1\right)}{y}\ge0\)

\(\Rightarrow M\ge\dfrac{\dfrac{4\left(y-1\right)^2}{y^2}+4}{y^2+1}=4.\dfrac{\left(y-1\right)^2+y^2}{y^2\left(y^2+1\right)}\)

\(\dfrac{M}{4}\ge\dfrac{2y^2-2y+1}{y^4+y^2}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{\left(2-y\right)\left(y^3+2y^2-3y+2\right)}{4\left(y^4+y^2\right)}+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(\Rightarrow M\ge1\)

Dấu "=" xảy ra khi \(y=2;x=1\)

tìm gtnn của biểu thức q=1/2(x^10/y^2 + y^10/x^2)+1/4(x^16 + y^16) - (1+ x^2y^2 )^2 

ai giúp mk vs