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a)
\(\dfrac{111}{37}=3< x< \dfrac{91}{13}=7\)
Vậy x = {4;5;6}
b)
\(-\dfrac{84}{14}=-6< 3x< \dfrac{108}{9}=12\Leftrightarrow-2< x< 4\)
Vậy x = {-1;0;1;2;3}
a, Ta có : \(\dfrac{111}{37}< x< \dfrac{91}{13}\)
\(\Rightarrow3< x< 7\)
Mà x là số nguyên .
\(\Rightarrow x\in\left\{4;5;6\right\}\)
b, Ta có : \(-\dfrac{84}{14.3}< x< \dfrac{108}{9.3}\)
\(\Rightarrow-2< x< 4\)
Mà x là số nguyên .
\(\Rightarrow x\in\left\{-1;0;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Theo đề bài 3<x<7.
=> x\(\in\){4;5;6}
b) tương tự như a)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 111/37 =3
91/13=7
-> x \(\in\left\{4,5,6\right\}\)
b) 84/14=6
108/9=12
mà 6<3x<12
-> 3x=9
->x=3.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2:để Z là số nguyên thì 3n-5 \(⋮\)n+4
\(\Rightarrow[(3n-5)-3(n+4)]⋮(n+4)\)
\(\Rightarrow(3n-5-3n-12)⋮(n+4)\)
\(\Rightarrow-17⋮n+4\)
\(\Rightarrow n+4\inƯ(17)\)={1;-1;17;-17}
\(\Rightarrow\)n\(\in\){-3;-5;13;-21}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>3<x<7
hay \(x\in\left\{4;5;6\right\}\)
b: =>-6<x<12
hay \(x\in\left\{-5;-4;-3;...;9;10;11\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)111/37<x<91/13
\(\Leftrightarrow\)\(\frac{1443}{481}\)<x<\(\frac{3367}{481}\)
vậy x\(\in\)\(\left\{\frac{1444}{481};......;\frac{3366}{481}\right\}\)
2)-84/14<x<108/9
\(\Leftrightarrow\)\(\frac{-801}{126}\)<x<\(\frac{432}{126}\)
vậy x\(\in\)\(\left\{\frac{-800}{126};.......;\frac{431}{126}\right\}\)
3)\(\frac{-x}{4}\)=\(\frac{-9}{x}\)\(\Rightarrow\)-x.x=(-x)2=4.(-9)=-36
\(\Rightarrow\)(-x)2=-36=(-6)2
vậy -x=6;x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...