Ta thấy: \(\left\{\begin{matrix}\left(x-y\right)^2\ge0\\\left(2x+3y-10\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow A=\left(x-y\right)^2+\left(2x+3y-10\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix}\left(x-y\right)^2=0\\\left(2x+3y-10\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x-y=0\\2x+3y-10=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=y\\2x+3y-10=0\end{matrix}\right.\)\(\Rightarrow2x+3x-10=0\)\(\Rightarrow5x=10\Rightarrow x=2\Rightarrow y=2\)

Vậy \(x=y=2\) để A đạt giá trị nhỏ nhất

Vì \(\left(x-y\right)^2\ge0;\left(2x+3y-10\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(2x+3y-10\right)^2\ge0\)

\(\Rightarrow A=\left(x-y\right)^2+\left(2x+3y-10\right)^2-2\ge-2\) có gtnn là - 2

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(2x+3y-10\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\2x+3y=10\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=2\end{cases}}}\)

Vật GTNN của A là - 2 <=> x = y = 2

\(\hept{\begin{cases}x-y=0\\2x+3y-10=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\6y=10\end{cases}\Rightarrow}x=y=\frac{10}{6}=\frac{5}{3}}\)

\(A=\left(x^2+xy+\dfrac{1}{4}y^2\right)-3\left(x+\dfrac{1}{2}y\right)+\dfrac{9}{4}+\left(\dfrac{3}{4}y^2+\dfrac{9}{2}y\right)-\dfrac{9}{4}\\ A=\left[\left(x+\dfrac{1}{2}y\right)^2-3\left(x+\dfrac{1}{2}y\right)+\dfrac{9}{4}\right]+\dfrac{3}{4}\left(y^2+6y+9\right)-9\\ A=\left(x+\dfrac{1}{2}y-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y+3\right)^2-9\ge9\\ A_{min}=9\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}y=\dfrac{3}{2}\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\Leftrightarrow2a-b=2\cdot3-3\left(-3\right)=12\)

Ta có: \(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\y=3x-2m+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)

Mặt khác: \(x^2+y^2=2m^2+2m+1=2\left(m^2+m+\dfrac{1}{2}\right)\)

                 \(=2\left(m^2+2\cdot m\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}\right)=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

 Dấu bằng xảy ra \(\Leftrightarrow m+\dfrac{1}{2}=0\Leftrightarrow m=-\dfrac{1}{2}\)

  Vậy ...

Ta có: P= \(x^2+xy+y^2-2x-3y+2010\)

\(\Leftrightarrow\) 4P= \(4\left(x^2+xy+y^2-2x-3y+2010\right)\)

= \(4x^2+4xy+4y^2-8x-12y+8040\)

= \(\left(4x^2+y^2+4+4xy-8x-8y\right)+3y^2-8y+8036\)

= \(\left(2x+y-2\right)^2+3y^2-8y+\dfrac{16}{3}-\dfrac{16}{3}+8036\)

= \(\left(2x+y-2\right)^2+3\left(y^2-\dfrac{8}{3}y+\dfrac{16}{9}\right)+\dfrac{24092}{3}\)

= \(\left(2x+y-2\right)^2+3\left(y-\dfrac{4}{3}\right)^2+\dfrac{24092}{3}\) \(\geq\) \(\dfrac{24092}{3}\)

\(\Rightarrow\) 4P \(\geq\) \(\dfrac{24092}{3}\) \(\Rightarrow\) P \(\geq\) \(\dfrac{6023}{3}\)

Dấu = xảy ra khi \(\begin{cases} (2x+y-2)^{2}=0\\ (y-\dfrac{4}{3})^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x+y-2=0\\ y-\dfrac{4}{3}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x=-(y-2)\\ y=\dfrac{4}{3} \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x=-(\dfrac{4}{3}-2)\\ y=\dfrac{4}{3} \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} x=\dfrac{1}{3}\\ y=\dfrac{4}{3} \end{cases} \)

Từ đó suy ra Min P= \(\dfrac{6023}{3}\) khi \(\begin{cases} x=\dfrac{1}{3}\\ y=\dfrac{4}{3} \end{cases} \)

Chúc bạn học tốt.

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