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b: (2x+1)(y-3)=-7
=>(2x+1;y-3) thuộc {(1;-7); (-7;1); (-1;7); (7;-1)}
=>(x,y) thuộc {(0;-4); (-4;4); (-1;10); (3;1)}
c: xy-x+y-1=3
=>(y-1)(x+1)=3
=>(x+1;y-1) thuộc {(1;3); (3;1); (-1;-3); (-3;-1)}
=>(x,y) thuộc {(0;4); (2;2); (-2;-2); (-4;0)}
d: =>x(y+3)+y+3=5
=>(x+1)(y+3)=5
=>(x+1;y+3) thuộc {(1;5); (5;1); (-1;-5); (-5;-1)}
=>(x,y) thuộc {(0;2); (4;-2); (-2;-8); (-6;-4)}
\(a,3x\left(y+1\right)+\left(y+1\right)=7\\ =>\left(3x+1\right)\left(y+1\right)=7\)
\(+,TH1:\left\{{}\begin{matrix}3x+1=1\\y+1=7\end{matrix}\right.=>\left\{{}\begin{matrix}x=0\\y=6\end{matrix}\right.\\ +,TH2:\left\{{}\begin{matrix}3x+1=7\\y+1=1\end{matrix}\right.=>\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ +,TH3:\left\{{}\begin{matrix}3x+1=\left(-1\right)\\y+1=\left(-7\right)\end{matrix}\right.=>\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=-8\end{matrix}\right.\\ +,TH4:\left\{{}\begin{matrix}3x+1=-7\\y+1=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=-\dfrac{8}{3}\\y=-2\end{matrix}\right.\)
a, 3x ( y+1) + y + 1 = 7
(y+1)(3x +1) =7
th1 : \(\left\{{}\begin{matrix}y+1=1\\3x+1=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)
th2: \(\left\{{}\begin{matrix}y+1=-1\\3x+1=-7\end{matrix}\right.\)=> x = -8/3 (loại)
th3: \(\left\{{}\begin{matrix}y+1=7\\3x+1=1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}y=6\\x=0\end{matrix}\right.\)
th 4 : \(\left\{{}\begin{matrix}y+1=-7\\3x+1=-1\end{matrix}\right.\)=> x=-2/3 (loại)
Vậy (x,y)= (2 ;0); (0; 6)
b, xy - x + 3y - 3 = 5
(x( y-1) + 3( y-1) = 5
(y-1)(x+3) = 5
th1: \(\left\{{}\begin{matrix}y-1=1\\x+3=5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)
th2: \(\left\{{}\begin{matrix}y-1=-1\\x+3=-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=-8\end{matrix}\right.\)
th3: \(\left\{{}\begin{matrix}y-1=5\\x+3=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=6\\x=-2\end{matrix}\right.\)
th4: \(\left\{{}\begin{matrix}y-1=-5\\x+3=-1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=-4\\x=-4\end{matrix}\right.\)
vậy (x, y) = ( 8; 2); ( -8; 0); (-2; 6); (-4; -4)
c, 2xy + x + y = 7 => y = \(\dfrac{7-x}{2x+1}\) ; y ϵ Z ⇔ 7-x ⋮ 2x+1
⇔ 14 - 2x ⋮ 2x + 1 ⇔ 15 - 2x - 1 ⋮ 2x + 1
th1 : 2x + 1 = -1=> x = -1; y = \(\dfrac{7-(-1)}{-1.2+1}\) = -8
th2: 2x+ 1 = 1=> x =0; y = 7
th3: 2x+1 = -3 => x = x=-2 => y = \(\dfrac{7-(-2)}{-2.2+1}\) = -3
th4: 2x+ 1 = 3 => x = 1 => y = \(\dfrac{7+1}{2.1+1}\) = 2
th5: 2x + 1 = -5 => x = -3=> y = \(\dfrac{7-(-3)}{-3.2+1}\) = -2
th6: 2x + 1 = 5 => x = 2; ; y = \(\dfrac{7-2}{2.2+1}\) =1
th7 : 2x + 1 = -15 => x = -8; y = \(\dfrac{7-(-8)}{-8.2+1}\) = -1
th8 : 2x+1 = 15 => x = 7; y = \(\dfrac{7-7}{2.7+1}\) = 0
kết luận
(x,y) = (-1; -8); (0 ;7); ( -2; -3) ; ( 1; 2); ( -3; -2); (2;1); (-8;-1);(7;0)
3xy−2x+5y=293xy−2x+5y=29
9xy−6x+15y=879xy−6x+15y=87
(9xy−6x)+(15y−10)=77(9xy−6x)+(15y−10)=77
3x(3y−2)+5(3y−2)=773x(3y−2)+5(3y−2)=77
(3y−2)(3x+5)=77(3y−2)(3x+5)=77
⇒(3y−2)⇒(3y−2) và (3x+5)(3x+5) là Ư(77)=±1,±7,±11,±77Ư(77)=±1,±7,±11,±77
Ta có bảng giá trị sau:
Do x,y∈Zx,y∈Z nên (x,y)∈{(−4;−3),(−2;−25),(2;3),(24;1)}
Ta có: 3x(y+1)+y+1=3x(y+1)+1(y+1)=(y+1)(3x+1)=7. Suy ra y+1 và 3x+1 là ước của 7. Suy ra x, y = (2, 0) hoặc (0, 6).
a) Thế x và y ta có:
\(-2.\left(-3\right)-5+11+3.\left(-3\right)\)
\(=6-5+11-9=3\)
b) Thế x và y ta có:
\(2.5-3.\left(-3\right)+5\left(5-\left(-3\right)\right)+15\)
\(=10+9+5\left(5+3\right)+15\)
\(=10+9+40+15=74\)
c) Thế x và y ta có:
\(4.\left(-3\right)-4\left(-3-2.5\right)-7\left(5-2\right)\)
\(=-12-4.\left(-13\right)-7.3\)
\(=-12+52-21=19\)
a) 0,5 x - \(\frac{2}{3}\)x = \(\frac{7}{12}\)
x . ( 0,5 - \(\frac{2}{3}\)) = \(\frac{7}{12}\)
x . \(\frac{-1}{6}\) = \(\frac{7}{12}\)
x = \(\frac{7}{12}\): \(\frac{-1}{6}\)
x = \(\frac{-7}{2}\)
b) x : \(\frac{25}{3}\) = -25
x = -25 . \(\frac{25}{3}\)
x = \(\frac{-625}{3}\)
c) 5,5 x = \(\frac{13}{15}\)
x = \(\frac{13}{15}\): 5,5
x = \(\frac{26}{165}\)
d) (\(\frac{3x}{7}\)+ 1 ) : ( -4) = \(\frac{-1}{28}\)
( \(\frac{3x}{7}\)+ 1 ) = \(\frac{-1}{28}\). ( -4)
\(\frac{3x}{7}\)+ 1 = \(\frac{1}{7}\)
\(\frac{3x}{7}\) = \(\frac{1}{7}\)- 1
\(\frac{3x}{7}\) = \(\frac{-6}{7}\)
=> x = -2
e) y + 30% y = -1,3
y + \(\frac{3}{10}\)y = -1,3
y . ( 1 + \(\frac{3}{10}\)) = -1,3
y . \(\frac{13}{10}\) = -1,3
y = -1,3 : \(\frac{13}{10}\)
y = -1
g) y - 25% y = \(\frac{1}{2}\)
y - \(\frac{1}{4}\)y = \(\frac{1}{2}\)
y . ( 1 - \(\frac{1}{4}\)) = \(\frac{1}{2}\)
y . \(\frac{3}{4}\) = \(\frac{1}{2}\)
y = \(\frac{1}{2}\): \(\frac{3}{4}\)
y = \(\frac{2}{3}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây huhu
=>(y+1)(3x+1)=7
=>\(\left(3x+1;y+1\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;16\right);\left(\dfrac{16}{3};0\right);\left(-\dfrac{2}{3};-18\right);\left(-6;-2\right)\right\}\)