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bn ơi bn lm đc bài này ko giúp mik vs
tìm x;y trong phương trình nghiệm nguyên sau:
a)x^2+y^2-2.(3x-5y)=11 b)x^2+4y^2=21+6x
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a.ta có \(\left(x+3\right)\left(y-7\right)=-21\Rightarrow y-7\in\left\{-3,-1\right\}\) ( do x+3>3 và 0>y-7>-7)
\(\Rightarrow\hept{\begin{cases}y=4\\x=4\end{cases}\text{ hoặc }}\hept{\begin{cases}y=6\\x=18\end{cases}}\)
c. \(\left(x-5\right)\left(y-5\right)=26=2\cdot13\Rightarrow x-5\in\left\{-2,-1,1,2,13,26\right\}\)
suy ra \(\left(x,y\right)\in\left\{\left(6,31\right);\left(31,6\right);\left(7,18\right);\left(18,7\right)\right\}\)
b.\(4xy+5y-14x=3\Leftrightarrow8xy+10y-28x=6\)
\(\Leftrightarrow\left(4x+5\right)\left(2y-7\right)=-29\)
mà 4x+5>5\(\Rightarrow4x+5=29\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)
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b ) x2 - 4x - 2y + xy + 1 = 0
( x2 - 4x + 4 ) - y ( 2 - x ) -3 = 0
( x - 2 )2 - y ( 2 - x ) = 3
( 2 - x ) ( 2 - x - y ) = 3
đến đây lập bảng tìm ra x,y
a) x2 + y2 + xy + 3x - 3y + 9 = 0
2x2 + 2y2 + 2xy + 6x - 6y + 18 = 0
( x2 + 2xy + y2 ) + ( x2 + 6x + 9 ) + ( y2 - 6y + 9 ) = 0
( x + y )2 + ( x + 3 )2 + ( y - 3 )2 = 0
\(\Rightarrow\)( x + y )2 = ( x + 3 )2 = ( y - 3 )2 = 0
\(\Rightarrow\)x = -3 ; y = 3
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\(pt\Leftrightarrow x\left(y-2\right)=-3y-1\)
\(\Leftrightarrow x=\frac{-3y-1}{y-2}=\frac{\left(-3y+6\right)-7}{y-2}=-3-\frac{7}{y-2}\)
Để \(x\inℤ\)thì \(\frac{7}{y-2}\inℤ\)
\(\Leftrightarrow y-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
lần lượt thay các giá trị của y-2 ta tìm dc các cặp nghiệm (x;y) là:
(-2; -5); (4; 1); (-10; 3); (-4; 9)