tìm số tự nhiên x ,biết : \(\dfrac{1}{3}\) < \(\dfrac{x}{2}\) < \(\dfrac{2}{3}\)
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\(\left\{\dfrac{-5< 0< -0,4}{x\in Z}\right\}\Rightarrow x\in\left\{-4;-3;-2;-1\right\}\)
1.
=2/5 x 12/3 + 2/5 x 15/3 + 2/5 x 1
= 2/5 x (12/3 + 15/3 + 1)
=2/5 x 1
=2/5
2.a=1;2
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)
=>-10<x<-13/7
hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
Từ gt ta có:
\(\dfrac{13}{3}.\left(-\dfrac{1}{3}\right)\le x\le\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\)
\(\Leftrightarrow\dfrac{-13}{9}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow\dfrac{-26}{18}\le x\le-\dfrac{11}{18}\)
Suy ra \(26\ge x\ge11\)
Vậy \(11\le x\le26\) ( x thuộc Z ) là các giá trị cần tìm
\(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{13}{3}.\dfrac{-1}{3}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)
\(\dfrac{-13}{9}\)\(\le x\le\)\(\dfrac{-11}{18}\)
\(\dfrac{-26}{18}\)\(\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x\in\left\{\dfrac{-12}{18};\dfrac{-13}{18};\dfrac{-14}{18};\dfrac{-15}{18};...;\dfrac{-24}{18};\dfrac{-25}{18}\right\}\)Tick hộ mình nha bạn
\(\dfrac{x+1}{2}=\dfrac{y-5}{3}=\dfrac{z-4}{4}=\dfrac{x+1+y-5-z+4}{2+3-4}\)
\(=\dfrac{7}{1}=7\)
\(\Rightarrow\left\{{}\begin{matrix}x=7.2-1=13\\y=7.3+5=26\\z=7.4+4=32\end{matrix}\right.\)
Áp dụng t/c dtsbn:
\(\dfrac{x+1}{2}=\dfrac{y-5}{3}=\dfrac{z-4}{4}=\dfrac{x+1+y-5-z+4}{2+3-4}=\dfrac{7+1+4-5}{1}=7\\ \Rightarrow\left\{{}\begin{matrix}x+1=14\\y-5=21\\z-4=28\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=13\\y=26\\z=32\end{matrix}\right.\)
\(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)
1313< x2x2< 2323
⇒⇒2626< x×36x×36< 4646
⇒⇒ 2 < x ××3 < 4
⇒⇒x ××3 = 3
x = 3 ÷÷3
x = 1