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

\(\Leftrightarrow\left(x+y\right)\left(x+2y-1\right)=-3\)

x² + 2y² +3xy - x - y + 3 = 0

(x² - y²) + (3y² + 3xy) - (x + y) = -3

(x - y)(x + y) + 3y(x + y) - (x + y) = -3

(x + y)(x + 2y -1) = -3 = 1.(-3) = (-1).3

(x;y)=(4;-3) (-6;5)

Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

1)<=> \(\left(x^2+2x+1\right)-\left(y^2+4x+4\right)-7=0\)

=> \(\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Rightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\left(x-y-1\right)\left(x+y+3\right)=7\)

vì x;y là số nguyên dương=> \(x;y\ge1\Rightarrow x+y+3\ge1+1+3=5\)

=> \(x+y+3=7\Rightarrow x+y=4\)

\(x-y-1=1\Rightarrow x-y=2\)

cộng hai phương trình ta có:

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

=> \(2x=6\)

=> x=3

=> y=4-3=1

2)<=> \(\left(x^2+xy+2x\right)+\left(2xy+2y^2+4y\right)+\left(x+y+2\right)=17\)

=> \(x\left(x+y+2\right)+2y\left(x+y+2\right)+1\left(x+y+2\right)=17\)

=> \(\left(x+y+2\right)\left(x+2y+1\right)=17\)

\(3xy+x+15y-44=0\)

\(3y\left(x+5\right)+\left(x+5\right)-49=0\)

\(\left(x+5\right)\left(3y+1\right)=49\)

Vì x;y là số nguyên \(\Rightarrow\hept{\begin{cases}x+5\in Z\\3y+1\in Z\end{cases}}\)

Có \(\left(x+5\right)\left(3y+1\right)=49\)

\(\Rightarrow\left(x+5\right)\left(3y+1\right)\in\text{Ư}\left(49\right)=\left\{\pm1;\pm7;\pm49\right\}\)

b tự lập bảng nhé~

không có phương trình bạn nhé

bạn ơi, xem lại đề ra 1 chút, hình như có câu sai đề thì phải

\(x^2+2y^2+3xy=5\)

=>\(x^2+xy+2xy+2y^2=5\)

=>\(x\left(x+y\right)+2y\left(x+y\right)=5\)

=>\(\left(x+y\right)\left(x+2y\right)=5\)

=>\(\left(x+y\right)\left(x+2y\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

TH1: \(\left\{{}\begin{matrix}x+y=1\\x+2y=5\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}-y=-4\\x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=1-y=1-4=-3\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=5\\x+2y=1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}-y=4\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-4\\x=5-y=5-\left(-4\right)=9\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}x+y=-1\\x+2y=-5\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}y=-4\\x=-5-2y=-5-2\cdot\left(-4\right)=-5+8=3\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}x+y=-5\\x+2y=-1\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}y=4\\x=-5-y=-5-4=-9\end{matrix}\right.\)

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k minh di mink giai cho de lam