\(A=x-y\)

+x<y => A<0

+ x>/ y =>\(A^2=\left(x-y\right)^2=\left(1.x+1.\left(-y\right)\right)^2\le\left(1^2+1^2\right)\left(x^2+y^2\right)=\frac{2.2025}{2}\)

 \(A\le45\)

=> Max \(A=45\) => x = -y  => 4 x² = 2025 => x =-y = 45/2

Vậy x =45/2 ; y =-45/2 

\(5x^2+5y^2+8xy-2x+2y+2=0\)

=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

=>x=1 và y=-1

\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)

A = x +y +1 => A - 1 = x +y.

Từ gt suy ra : (A -1)² + 7(A -1) + y² + 10 = 0 => A² + 5A + 4 + y² = 0 => A² + 5A + 4 = - y² <= 0. Dấu = xảy ra khi y = 0

=> (A +1)(A +4) <= 0 => - 1 <= A <= -4

A = -1 <=> y = 0 và x + y = -1 => y = 0 và x = -1

A = -4 <=> y =0 và x + y = -4 => y = 0 và x = -4

Vậy minA = -1 khi x = -1, y = 0

maxA = -4 khi x = -4, y = 0

\(4=2^x+2^y\ge2\sqrt{2^{x+y}}\Rightarrow2^{x+y}\le4\Rightarrow x+y\le2\)

\(\Rightarrow xy\le1\)

\(P=4x^2y^2+2x^3+2y^3+10xy\)

\(P=4x^2y^2+10xy+2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(P\le4x^2y^2+10xy+4\left(4-3xy\right)=4x^2y^2-2xy+16\)

Đặt \(xy=t\Rightarrow0< t\le1\)

Xét hàm \(f\left(t\right)=4t^2-2t+16\) trên \((0;1]\)

\(\Rightarrow...\)

Đặt \(\left\{{}\begin{matrix}x-4=a\\y-3=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=5\)

\(Q=\sqrt{\left(a+5\right)^2+b^2}+\sqrt{\left(a+3\right)^2+\left(b+4\right)^2}\)

\(\Rightarrow Q\le\sqrt{2\left[\left(a+5\right)^2+b^2+\left(a+3\right)^2+\left(b+4\right)^2\right]}\) (Bunhiacopxki)

\(\Rightarrow Q\le\sqrt{4\left(a^2+8a+b^2+4b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2.4a+b^2+2.2b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2\left(a^2+4\right)+b^2+2\left(b^2+1\right)+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(3a^2+3b^2+35\right)}\le\sqrt{4\left(3.5+35\right)}=10\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=4\end{matrix}\right.\)