BT4: Tìm số nguyên x, biết
1) \(-4\dfrac{3}{5}.2\dfrac{4}{43}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
K
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24 tháng 9 2022
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)
=>-10<=x<=-13/7
hay \(x\in\left\{-10;-9;...;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=11/18
hay \(x\in\left\{-1;0\right\}\)
21 tháng 5 2022
\(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{3}\cdot\dfrac{4-6-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< =x< =\dfrac{-2}{3}\cdot\dfrac{-11}{12}=\dfrac{22}{36}=\dfrac{11}{18}\)
mà x là số nguyên
nên \(x\in\left\{-1;0;1\right\}\)
N
27 tháng 6 2018
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 ...
TH
27 tháng 6 2018
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!!!
30 tháng 5 2017
4\(\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\)\(\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{-13}{9}\le x\le\dfrac{-11}{12}\)
\(\dfrac{-468}{36}\le\dfrac{36.x}{36}\le\dfrac{-396}{36}\)
\(=>36.x\in\left\{-467;-466;-465;-464;...;-398;-397\right\}\)
\(=>x=-12\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 9 2023
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
23 tháng 9 2023
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
QD
2
12 tháng 9 2021
\(-4\dfrac{3}{5}\cdot2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}\cdot1\dfrac{6}{15}\)
\(\Leftrightarrow-10\le x\le-\dfrac{91}{25}\)
\(\Leftrightarrow x\in\left\{-9;-8;-7;-6;-5;-4\right\}\)
12 tháng 9 2021
\(\Rightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}\le x\le-\dfrac{13}{5}\cdot\dfrac{7}{5}\\ \Rightarrow-10\le x\le-\dfrac{91}{25}\\ \Rightarrow-\dfrac{2500}{25}\le x\le-\dfrac{91}{25}\\ \Rightarrow x\in\left\{-\dfrac{2499}{25};-\dfrac{2498}{25};...;-\dfrac{92}{25};-\dfrac{91}{25}\right\}\)
\(\Leftrightarrow-\dfrac{43}{5}\cdot\dfrac{90}{43}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)
=>-18<=x<=-13/7
mà x là số nguyên
nên \(x\in\left\{-18;-17;...;-2\right\}\)