BT1: Tìm số nguyên x, biết:
1) \(-3+\dfrac{1}{2}< x< \dfrac{7}{2}+\dfrac{1}{3}\)
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DH
24 tháng 7 2017
\(-\dfrac{2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{7}{6}\)
\(\Rightarrow\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\)
\(\Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\)
\(\Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\Rightarrow x=\dfrac{147}{20}\)
Chúc bạn học tốt!!!
21 tháng 1 2022
\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)
28 tháng 12 2021
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
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
G
18 tháng 10 2017
Giải:
\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
18 tháng 10 2017
\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\\\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)=\left(\pm\dfrac{1}{2}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=\dfrac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{21}{10}\\x=\dfrac{9}{10}\end{matrix}\right. \)
Vậy \(x=\dfrac{21}{10}\) hoặc \(x=\dfrac{9}{10}\)
DT
Đỗ Thanh Hải
CTVVIP
8 tháng 8 2017
\(x+\left|\dfrac{1}{2}-\dfrac{1}{3}\right|=\left|\dfrac{-2}{3}-\dfrac{1}{4}\right|\)
\(x+\left|\dfrac{1}{6}\right|=\left|\dfrac{-11}{12}\right|\)
\(x+\dfrac{1}{6}=\dfrac{11}{12}\)
\(x=\dfrac{11}{12}-\dfrac{1}{6}\)
\(x=\dfrac{3}{4}\)
Vậy ...
1) \(-3+\dfrac{1}{2}< x< \dfrac{7}{2}+\dfrac{1}{3}\)
\(\Rightarrow-\dfrac{5}{2}< x< \dfrac{23}{6}\)
\(\Rightarrow-2,5< x< 3,8\left(3\right)\)
Mà x là số nguyên nên:
\(x\in\left\{-2;-1;0;1;2;3\right\}\)
\(-3+\dfrac{1}{2}< x< \dfrac{7}{2}+\dfrac{1}{3}\)
\(\Rightarrow-2\dfrac{1}{2}< x< 3\dfrac{5}{6}\)
Vì \(x\in Z\) nên \(x\in\left\{-2;-1;0;1;2;3\right\}\)
Vậy...............
Chúc bạn học tốt!!!