a: Thay x=2 và y=y vào hệ, ta được:

my+2=2 và 2m-2y=1

=>my=0 và 2m-2y=1

=>\(m\in\varnothing\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\m\left(2-my\right)-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\2m-m^2y-2y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\y\left(-m^2-2\right)=1-2m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2m-1}{m^2+2}\\x=2-\dfrac{2m^2-m}{m^2+2}=\dfrac{2m^2+4-2m^2+m}{m^2+2}=\dfrac{m+4}{m^2+2}\end{matrix}\right.\)

Để \(S=2x-y=\dfrac{2m+8-2m+1}{m^2+2}=\dfrac{7}{m^2+2}_{MAX}\) thì m^2+2 min

=>m=0

hình như thiếu đề

còn câu a, mà mình giải rồi nhé bạn 

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m+1}{m^2}\ne\dfrac{-2}{-1}=2\)

=>\(2m^2\ne m+1\)

=>\(2m^2-m-1\ne0\)

=>\(\left(m-1\right)\left(2m+1\right)\ne0\)

=>\(m\notin\left\{1;-\dfrac{1}{2}\right\}\)

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

=>\(\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\2m^2\cdot x-2y=2m^2+4m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(2m^2-m-1\right)=2m^2+4m-m+1\\\left(m+1\right)x-2y=m-1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\2y=\left(m+1\right)x-\left(m-1\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\2y=\dfrac{m^2+2m+1-\left(m-1\right)^2}{m-1}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\y=\dfrac{m^2+2m+1-m^2+2m-1}{2m-2}=\dfrac{4m}{2m-2}=\dfrac{2m}{m-1}\end{matrix}\right.\)

Để x,y đều nguyên thì \(\left\{{}\begin{matrix}m+1⋮m-1\\2m⋮m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m-1+2⋮m-1\\2m-2+2⋮m-1\end{matrix}\right.\)

=>\(2⋮m-1\)

=>\(m-1\in\left\{1;-1;2;-2\right\}\)

=>\(m\in\left\{2;0;3;-1\right\}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\2m^2x-2y=2m^2+4m\end{matrix}\right.\)

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

Pt có nghiệm duy nhất khi \(2m^2-m-1\ne0\Rightarrow m\ne\left\{1;-\dfrac{1}{2}\right\}\)

Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{2m^2-2m-1}{2m^2+3m+1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{\left(m+1\right)\left(2m+1\right)}=\dfrac{m-1}{m+1}\\y=m^2x-m^2-2m=\dfrac{-4m^2-2m}{m+1}\end{matrix}\right.\)

Để x nguyên \(\Rightarrow\dfrac{m-1}{m+1}\in Z\Rightarrow1-\dfrac{2}{m+1}\in Z\)

\(\Rightarrow\dfrac{2}{m+1}\in Z\)

\(\Rightarrow m+1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow m=\left\{-3;-2;0;1\right\}\)

Thay vào y thấy đều thỏa mãn y nguyên.

Vậy ...

\(\left\{{}\begin{matrix}\sqrt{2x}+\sqrt{3-y}=m\left(1\right)\\\sqrt{2y}+\sqrt{3-x}=m\left(2\right)\end{matrix}\right.\) \(\left(0\le x,y\le3\right)\)

\(\left(1\right)-\left(2\right)\Leftrightarrow\sqrt{2x}-\sqrt{2y}+\sqrt{3-y}-\sqrt{3-x}=0\)

\(\Leftrightarrow\dfrac{2x-2y}{\sqrt{2x}+\sqrt{2y}}+\dfrac{3-y-3+x}{\sqrt{3-y}+\sqrt{3-x}}=0\Leftrightarrow\left(x-y\right)\left(\dfrac{2}{\sqrt{2x}+\sqrt{2y}}+\dfrac{1}{\sqrt{3-y}+\sqrt{3-x}}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\\dfrac{2}{\sqrt{2x}+\sqrt{2y}}+\dfrac{1}{\sqrt{3-y}+\sqrt{3-x}}=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\left(1\right)và\left(3\right)\Rightarrow\sqrt{2x}+\sqrt{3-x}=m\)

\(m^2=x+3+2\sqrt{2x\left(3-x\right)}\ge3\Leftrightarrow\left[{}\begin{matrix}m\ge\sqrt{3}\\m\le-\sqrt{3}\end{matrix}\right.\)\(\left(4\right)\)

\(m\le\sqrt{3\left(x+3-x\right)}=3\left(5\right)\)

\(\left(4\right)\left(5\right)\Rightarrow\sqrt{3}\le m\le3\Rightarrow m=\left\{2;3\right\}\)

Trừ vế cho vế:

\(\sqrt{2x}-\sqrt{2y}+\sqrt{3-y}-\sqrt{3-x}=0\)

\(\Rightarrow\dfrac{\sqrt{2}\left(x-y\right)}{\sqrt{x}+\sqrt{y}}+\dfrac{x-y}{\sqrt{3-y}+\sqrt{3-x}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{3-y}+\sqrt{3-x}}\right)=0\)

\(\Leftrightarrow x=y\)

Thế vào pt đầu:

\(\sqrt{2x}+\sqrt{3-x}=m\)

Ta có: \(\sqrt{2.x}+\sqrt{1.\left(3-x\right)}\le\sqrt{\left(2+1\right)\left(x+3-x\right)}=3\)

\(\sqrt{2x}+\sqrt{3-x}=\sqrt{x}+\sqrt{3-x}+\left(\sqrt{2}-1\right)\sqrt{x}\ge\sqrt{x+3-x}+\left(\sqrt{2}-1\right)\sqrt{x}\ge\sqrt{3}\)

\(\Rightarrow\sqrt{3}\le m\le3\Rightarrow m=\left\{2;3\right\}\)

heo my

Hệ phương trình tương đương:\(\hept{\begin{cases}2x+y=5m-1\\2x-4y=4\end{cases}}\)

Trừ theo vế ta được:\(2x-2x+y-\left(-4y\right)=5m-1-4\)

Hay: \(5y=5\left(m-1\right)\)

Tương đương: \(y=m-1\)

Thay \(y=m-1\)vào 1 trong 2 đẳng thức trên ta được \(x=2m\)

Vậy hệ phương trình luôn có nghiệm duy nhất \(\hept{\begin{cases}x=2m\\y=m-1\end{cases}}\)

Để \(x>0,y>0\)thì \(\hept{\begin{cases}2m>0\\m-1>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}m>0\\m>1\end{cases}}\)\(\Rightarrow m>1\)

Vậy \(m>1\)thì hệ phương trình có nghiệm \(x>0,y>0\)