a: =>-12<x<2y<-9

=>x=-11; y=-5

b: =>-7<3(x-1)<8

\(\Leftrightarrow3\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)

\(\Leftrightarrow x-1\in\left\{2;1;0;-1;-2\right\}\)

hay \(x\in\left\{3;2;1;0;-1\right\}\)

Ta có: x(x+y)+y(x+y)=36+64

=>(x+y)(x+y)=100

=>(x+y)^2=10^2=(-10)^2

Với x+y=10,ta có: x.(x+y)=36=>x.10=36=>x=36:10=3,6

y(x+y)=64=>y.10=64=>y=64:10=6,4

Với x+y=-10,ta có: x(x+y)=36=>x.(-10)=36=>x=36:(-10)=-3,6

y(x+y)=64=>y.(-10)=64=>y=64:(-10)=-6,4

Vậy (x;y) \(\in\){(3,6;6,4);(-3,6;-6,4)}

x,y=3,-3,6,-6,4,-4

A

Chọn A

x = 48

y= 12                       

\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4;\frac{x}{12}=4\Rightarrow x=4.12\Rightarrow x=48;\frac{y}{3}=4\Rightarrow y=4.3\Rightarrow y=12\)

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a.\(y=\dfrac{k}{x}\Rightarrow k=y.x=12.9=108\)

b.khi x=4 thì \(y=\dfrac{k}{x}=\dfrac{108}{4}=27\)

c.khi y=36 thì \(x=\dfrac{k}{y}=\dfrac{108}{36}=3\)