Áp dụng BĐT Bunhiaskopski:

\(A^2=\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)

\(A^2\le25\Rightarrow-5\le A\le5\)

Max:Dấu ''='' xảy ra khi x=y=1

Min:Dấu ''='' xảy ra khi x=y=-1

Hok bít đúng hok nữa, sai thôi nha

Áp dụng bđt \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

\(\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5^2\)

\(\Rightarrow-5\le2x+3y\le5\)

Dấu bằng xảy ra khi \(\frac{a}{x}=\frac{b}{y}\)hay \(\frac{\sqrt{2}x}{\sqrt{3}y}=\frac{\sqrt{2}}{\sqrt{3}}\Leftrightarrow x=y\)

Vậy \(A\text{ min }=-5\Leftrightarrow x=y=-1\)

\(A\text{ max }=5\Leftrightarrow x=y=1\)

\(A^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\)

\(\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)

\(\Rightarrow-5\le A\le5\)

\(A_{max}=5\) khi \(x=y=1\)

\(A_{min}=-5\) khi \(x=y=-1\)

\(A=\sqrt{1-x}+\sqrt{x+1}\)

\(A^2=\left(\sqrt{1-x}\cdot1+\sqrt{x+1}\cdot1\right)^2\)

Áp dụng BĐT Bunhiacospki ta có:
\(A^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)\)

\(A^2\le4\)

\(A\le2\)

\(A_{max}=2\Leftrightarrow x=0\)

E ms tìm dc MAX thôi ah

ĐKXĐ: ....

a/ \(A\le\sqrt{2\left(1-x+1+x\right)}=2\Rightarrow A_{max}=2\) khi \(x=0\)

\(A\ge\sqrt{1-x+1+x}=\sqrt{2}\Rightarrow A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

b/ \(B\le\sqrt{2\left(x-2+6-x\right)}=2\sqrt{2}\Rightarrow B_{max}=2\sqrt{2}\) khi \(x=4\)

\(B\ge\sqrt{x-2+6-x}=2\Rightarrow B_{min}=2\) khi \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

c/ \(A^2=\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\)

\(\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)

\(\Rightarrow-5\le A\le5\)

\(A_{max}=5\) khi \(x=y=1\)

\(A_{min}=-5\) khi \(x=y=-1\)

GTLN:

Áp dụng BĐT \(a^2+b^2\ge2ab\)

\(\Rightarrow x^2+1\ge2x\Rightarrow2x^2\ge4x-2\)

\(y^2+1\ge2y\Rightarrow3y^2\ge6y-3\)

\(\Rightarrow2x^2+3y^2\ge2\left(2x+3y\right)-5\)

mà \(2x^2+3y^2\le5\)

\(\Rightarrow2\left(2x+3y\right)-5\le5\Rightarrow2x+3y\le5\)

Vậy Max A = 5 khi x = y = 1

có làm thì mới có ăn ko làm mà đòi có ăn thì ăn đồng bằng ăn cát

Ta thấy \(8x^3+27y^3\)

\(=\left(2x\right)^3+\left(3y\right)^3\)

\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

\(=4x^2-6xy+9y^2\)

Thế thì \(A=6x^2-6xy+18y^2+5\)

Rồi lại thay \(x=\dfrac{1-3y}{2}\) vào A thôi.