Tìm số nguyên x biết:
a) \(-4\dfrac{3}{5}\) x \(2\dfrac{4}{3}\) \(\le\) x \(\le\) \(-2\dfrac{3}{5}\) : \(1\dfrac{6}{15}\)
b) \(-4\dfrac{1}{3}\) (\(\dfrac{1}{2}-\dfrac{1}{6}\) ) \(\le\) x \(\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{7}\right)\)
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}\dfrac{x+6}{30}\)
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
mk quên mất, còn câu b/ nữa, câu c thì bạn Nháy giải rồi nên mk khỏi giải nha.
b/\(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\le x\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{7}\right)\)
\(\Rightarrow\)\(\dfrac{-13}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\le x\le\dfrac{2}{3}\left(\dfrac{14}{42}-\dfrac{21}{42}-\dfrac{18}{42}\right)\)
\(\Rightarrow\dfrac{-13}{3}.\dfrac{1}{3}\le x\le\dfrac{2}{3}.\dfrac{-25}{42}\)
\(\Rightarrow\dfrac{-13}{9}\le x\le\dfrac{-25}{63}\)
\(\Rightarrow-1,\left(4\right)\le x\le0,\left(396825\right)\)
Vì x\(\in Z\) nên x \(\in\left\{-1;0\right\}\)
Mk chỉ nghĩ z thôi chứ có j thì góp ý thêm nha.
