Tìm các số nguyên x,y biết:
a) x/y=2/5
b) x/3=y/7
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a: (x-2)(y-3)=5
=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)
b: (2x-1)*(y-4)=-11
=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)
=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)
c: xy-2x+y=3
=>\(x\left(y-2\right)+y-2=1\)
=>\(\left(x+1\right)\left(y-2\right)=1\)
=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)
=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)
`A)2/3=x/60`
`=>40/60=x/60`
`=>x=40`
`B)-1/2=y/18`
`=>-9/18=y/18`
`=>y=-9`
`C)3/x=y/35=-36/84`
Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`
`=>3/x=-3/7`
`=>x=-7`
`y/35=-3/7=-15/35`
`=>y=-15`
`D)7/x=y/27=-42/54`
Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`
`=>7/x=-7/9`
`=>x=-9`
`y/27=-7/9=-21/27`
`=>y=-21`
b: =>3x+9=0 và y^2-9=0 và x+y=0
=>x=-3; y=3
a: (2x-5)(y+3)=-22
mà x,y là số nguyên
nên \(\left(2x-5;y+3\right)\in\left\{\left(1;-22\right);\left(11;-2\right);\left(-1;22\right);\left(-11;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;-25\right);\left(8;-5\right);\left(2;19\right);\left(-3;-1\right)\right\}\)
a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
x+1 | 1 | -1 | 5 | -5 |
y-2 | 5 | -5 | 1 | -1 |
x | 0 | -2 | 4 | -6 |
y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
x-5 | 1 | -1 | 7 | -7 |
y+4 | -7 | 7 | -1 | 1 |
x | 6 | 4 | 12 | -2 |
y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)
Lời giải:
a. $\frac{x}{7}=\frac{6}{21}$
$x=\frac{6}{21}.7$
$x=2$
b.
$\frac{-5}{y}=\frac{20}{28}$
$y=-5:\frac{20}{28}$
$y=-7$
c.
$\frac{-4}{8}=\frac{-7}{y}$
$y=-7:\frac{-4}{8}$
$y=14$
a, \(\dfrac{x}{7}=\dfrac{6}{21}\Leftrightarrow\dfrac{3x}{21}=\dfrac{6}{21}\Rightarrow x=2\)
b, \(\dfrac{-5}{y}=\dfrac{20}{28}\Leftrightarrow\dfrac{20}{-4y}=\dfrac{20}{28}\Leftrightarrow y=-7\)
c, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)
a: =>y/15=-2/3
hay y=-10
b: 2/x=x/18
nên \(x^2=36\)
hay \(x\in\left\{6;-6\right\}\)
c: x/9=16/x
nên \(x^2=144\)
hay \(x\in\left\{12;-12\right\}\)
a: =>x/-3=3
hay x=-9
b: =>x/9=-1/9
hay x=-1
c: =>x+1/5=-1/3
hay x=-8/15
d: =>-7/x=-7/9
hay x=9
a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)
b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)
c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)
d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)