\(a,xy-x-y=2\\ x\left(y-1\right)-y=2\\ x\left(y-1\right)-y+1=2+1\\ x\left(y-1\right)-\left(y-1\right)=3\\ \left(y-1\right)\left(x-1\right)=3\\ Th1:x-1=-1=>x=0\\ y-1=-3=>y=-2\\ Th2:x-1=-3 =>x=-2\\ y-1=-1=> y=0\\ Th3:x-1=3=> x=4\\ y-1=1=>y=2\\ Th4:x-1=1=>x=2\\ y-1=3=>y=4\)

Vậy......

\(b,2x^2+3xy-2y^2=7\\ 2x^2+\left(4xy-xy\right)-2y^2=7\\ x\left(2x-y\right)+2y\left(2x-y\right)=7\\ \left(2x-y\right)\cdot\left(x+2y\right)=7\)

Nếu 2x-y=1; x+2y = 7

=> 2(2x-y) + x + 2y = 9

=> 4x - 2y + x +2y = 9

=> (4x+x) + (2y-2y) = 9

=> 5x + 0 = 9 

=> x = 9/5 (ktm)

Nếu 2x-y=7; x+2y = 1

=> 2(2x-y) + x+ 2y = 15

=> 4x - 2y + x +2y =15

=> (4x +x)+ (2y-2y) =15

=> 5x +0 =15

=> x= 3 (tm)

=> y= -1 (Tm)

Nếu  2x-y=-7; x+2y = -1

=> 2(2x-y) + x+ 2y = -15

=> 4x - 2y + x +2y =-15

=> (4x +x)+ (2y-2y) =-15

=> 5x +0 =-15

=> x= -3 (tm)

=> y= 1 (tm)

Nếu 2x-y=-1 ; x+2y = -7

=> 2(2x-y) + x+ 2y = -9

=> 4x - 2y + x +2y = -9

=> (4x +x)+ (2y-2y) =-9

=> 5x +0 =-9

=> x= -9/5 (ktm)

=> y= -1

Vậy.........

a) \(2x^2-3xy-2y^2=2\)

\(\Rightarrow2x^2+xy-4xy-2y^2=2\)

\(\Rightarrow x\left(2x+y\right)-2y\left(2x+y\right)=2\)

\(\Rightarrow\left(2x+y\right)\left(x-2y\right)=2\)

\(\Rightarrow\left(2x+y\right);\left(x-2y\right)\in\left\{-1;1;-2;2\right\}\)

Ta giải các hệ phương trình sau với x;y nguyên 

1) \(\left\{{}\begin{matrix}2x+y=-1\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}2x+y=1\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}4x+2y=2\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}5x=4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}2x+y=-2\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-4\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

4)  \(\left\{{}\begin{matrix}2x+y=2\\x-2y=1\end{matrix}\right.\) \(\left\{{}\begin{matrix}4x+2y=4\\x-2y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right)\right\}\)

b) \(xy-y+x=9\)

\(\Rightarrow y\left(x-1\right)+x-1+1=9\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=8\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;-9\right);\left(2;7\right);\left(-1;-5\right);\left(3;3\right);\left(-3;-3\right);\left(5;1\right);\left(-7;-2\right);\left(9;0\right)\right\}\)

\(x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=100\)

Ta có: \(2x^2+xy+2y^2=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x^2+2xy+y^2\right)=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\)

Theo BĐT Bunhacopxky: \(\left(x^2+y^2\right)\left(1+1\right)\ge\left(x+y\right)^2\Rightarrow\dfrac{3}{2}\left(x^2+y^2\right)\ge\dfrac{3}{4}\left(x+y\right)^2\\ \Rightarrow2x^2+xy+2y^2=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{5}{4}\left(x+y\right)^2\\ \Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

Chứng minh tương tự:

\(\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)\\ \sqrt{2z^2+xz+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)

Cộng vế theo vế, ta được: \(P\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\cdot1=\sqrt{5}\)

Dấu "=" \(\Leftrightarrow x=y=z=\dfrac{1}{3}\) 

Bạn tham khảo nhé

https://hoc24.vn/cau-hoi/cho-cac-so-duong-xyz-thoa-man-xyz1cmrcan2x2xy2y2can2y2yz2z2can2z2zx2x2can5.182722154737

a) \(P=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)

\(P=3\left(x+y\right)^2-2.5-100\)

\(P=3.5^2-110\)

\(P=-35\)

b) \(Q=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3.5+10\)

\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+25\)

\(Q=5^3-2.5^2+25\)

\(Q=100\)

\(y\in\left(-\infty;\infty\right)\)

\(-2y^2-3xy-2y+2x^2+6x=1\)

\(-2y^2-3xy-2y-2x^2+6x-1=0\)

\(-2y^2-\left(3x+2\right)y+2x^2+6x-1=0\)

\(y=\frac{\sqrt{25x^2+60x-4-3x-2}}{4}\)

\(y=-\frac{\sqrt{25x^2+60x-4+3x+2}}{4}\)

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