\(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay x - y = 7

\(\Rightarrow A=49+14+37=100\)

Vậy A = 100 khi x - y = 7

B= \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\)

ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\)

=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

=>\(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\frac{9}{16}\)

=> min B=9/16 kh x=-1/2

C= \(x^2-2xy+y^2+1\)= \(\left(x-y\right)^2+1\)

ta có \(\left(x-y\right)^2\ge0\)=>\(\left(x-y\right)^2+1\ge1\)

=> Min C=1 khi x=y

2xy-10x+y=17

\(\Rightarrow2x\cdot\left(y-5\right)+y=17\)

\(\Rightarrow2x\left(y-5\right)+y-5=17-5\)

\(\Rightarrow2x\left(y-5\right)+\left(y-5\right)=12\)

\(\Rightarrow\)(2x+1)(y-5)=12

Xong xét từng giá trị là ra

\(A=\dfrac{-\left(x^2+2xy+y^2\right)+4x^2+4xy+y^2}{x^2+2xy+y^2}=-1+\left(\dfrac{2x+y}{x+y}\right)^2\ge-1\)

\(A_{min}=-1\) khi \(2x+y=0\)

x²+2xy+y²=9

=>(x²+xy)+(xy+y²)=9

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

=>(x+y)(x+y)=3.3

=>x+y=3

x²-2xy+y²=1

=>(x²-xy)+(y²-xy)=1

=>x(x-y)+y(y-x)=1

=>x(x-y)-y(x-y)=1

=>(x-y)(x-y)=1.1

=>x-y=1

x+y+x-y=3+1

=>2x=4

=>x=2

=>y=2-1

=>y=1

vậy x=2 và y=1

a)ta có :2xy-6=4x-y => 2xy-6-4x+y=0 => 2*(2xy-6-4x+y)=2*0               =>4xy-12-8x+2y=0 => 2x2y-4-8-8x+2y=0 => 2x2y-4-8x+2y=8                 =>(2x2y+2y)-(8x+4)=8 =>2y(2x+1)-4(2x+1)=8 => (2y-4)(2x+1)=8           Ta có bảng sau :

Vậy các cặp nghiệm x,y thỏa mãn là (0;6) và (-1;-2)

\(2xy^2+x+y-1=x^2+2y^2+xy\\\Leftrightarrow 2xy^2+x+y-1-x^2-2y^2-xy=0\\\Leftrightarrow(2xy^2-2y^2)-(xy-y)-(x^2-x)=1\\\Leftrightarrow2y^2(x-1)-y(x-1)-x(x-1)=1\\\Leftrightarrow(x-1)(2y^2-y-x)=1\)

Vì \(x,y\) nguyên \(\Rightarrow x-1;2y^2-y-x\) có giá trị nguyên

Mà: \(\left(x-1\right)\left(2y^2-y-x\right)=1\)

Do đó ta có các trường hợp xảy ra là:

\(+,\left\{{}\begin{matrix}x-1=1\\2y^2-y-x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\\left(2y-3\right)\left(y+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y\in\left\{\dfrac{3}{2};-1\right\}\end{matrix}\right.\)

Mà \(x,y\) nguyên nên: \(x=2;y=-1\)

\(+,\left\{{}\begin{matrix}x-1=-1\\2y^2-y-x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2\left(y-\dfrac{1}{4}\right)^2+\dfrac{7}{8}=0\left(\text{vô lí}\right)\end{matrix}\right.\)

Vậy \(x=2;y=-1\) là các giá trị cần tìm.

\(\text{#}Toru\)

2xy-x-y=2
=>2xy-y-x=2
=>y(2x-1)-(2x-1)+x-1=2
=>(2x-1)(y-1)+x=3
=>(2x-1)(y-1) và x là Ư(3)={1;-1;3;-3}
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