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 \(x=y=-1\)

Ta có 

\(1A^2=\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)\)

\(\le5.5=25\)

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

Vậy GTNN là - 5 đạt được khi x = y = - 1

tuong Min=5 chu

thay vao cx thay no dung hinh nhu Min=-5 that

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

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

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

Biết x^2+y^2=52 
tìm GTLN,GTNN của A=2x+3y 

áp dụng H) có: 
A² = (2x+3y)² ≤ (4 + 9)(x² + y²) = 13.52 = 676 
=> - 26 ≤ A ≤ 26 
Amin = - 26 ; A max = 26 đạt được khi: 
x/y = 2/3 <=> x = 2y/3 kết hợp x² + y² = 52 => y² + 4y²/9 = 52 <=> y= ± 6 , x = ± 4

Câu 3 là (1+1/x)(1+1/y) nha

Mà ko cần làm câu này đâu giúp mình 2 câu 1 và 2 thôi nhá

\(2x+3y=1\Rightarrow y=\frac{1-2x}{3}\)

Do \(x;y\ge0\Rightarrow0\le x\le\frac{1}{2}\)

\(A=x^2+3\left(\frac{1-2x}{3}\right)^2=x^2+\frac{1}{3}\left(4x^2-4x+1\right)=\frac{7}{3}x^2-\frac{4}{3}x+\frac{1}{3}\)

\(A=\frac{7}{3}\left(x-\frac{2}{7}\right)^2+\frac{1}{7}\ge\frac{1}{7}\)

\(\Rightarrow A_{min}=\frac{1}{7}\) khi \(x=\frac{2}{7};y=\frac{1}{7}\)

Mặt khác \(A=\frac{1}{3}x\left(7x-4\right)+\frac{1}{3}\)

Do \(x\le\frac{1}{2}\Rightarrow7x-4< 0\Rightarrow x\left(7x-4\right)\le0\)

\(\Rightarrow A\le\frac{1}{3}\Rightarrow A_{max}=\frac{1}{3}\) khi \(x=0;y=\frac{1}{3}\)

Câu 2:

\(A-4=2x+3y\Rightarrow\left(A-4\right)^2=\left(2x+3y\right)^2\)

\(\left(A-4\right)^2\le\left(2^2+3^2\right)\left(x^2+y^2\right)=676\)

\(\Rightarrow-26\le A-4\le26\)

\(\Rightarrow-22\le A\le30\)

\(A_{max}=30\) khi \(\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

\(A_{min}=-22\) khi \(\left\{{}\begin{matrix}x=-4\\y=-6\end{matrix}\right.\)