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1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
a) \(\left(x-\frac{1}{2}\right).\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\left(x-\frac{1}{2}\right).\frac{1}{3}=9\frac{5}{7}-\frac{5}{7}\)
\(x-\frac{1}{2}=9:\frac{1}{3}\)
\(x=27+\frac{1}{2}\)
\(x=\frac{55}{2}\)
b)\(\frac{1}{2}.x+150\%.x=2014\)
\(\left(\frac{1}{2}+150\%\right)x=2014\)
\(2x=2014\)
\(x=2014:2\)
\(x=1007\)
c) \(2-\left|\frac{3}{4}\right|=\frac{7}{2}\)???????
và bài này nữa
tính nhanh
a, \(5\frac{7}{5}\cdot4\frac{2}{7}+5\frac{5}{7}\cdot4\frac{2}{7}\)
b, \(b,\frac{-7}{11}\cdot\frac{4}{19}+\frac{-7}{19}\cdot\frac{7}{11}+2\frac{7}{19}\)
cảm ơn giúp nhiều
1a)
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}=>\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)
\(\left(\frac{4}{6}-\frac{9}{6}\right)x=\frac{5}{12}=>\frac{-5}{6}x=\frac{5}{12}\)
\(x=\frac{5}{12}:\frac{-5}{6}=>x=\frac{5}{12}.\frac{-6}{5}=>x=\frac{-1}{2}\)
b)\(\frac{-2}{3}x+\frac{1}{5}=\frac{3}{10}=>\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}\)
\(\frac{-2}{3}x=\frac{3}{10}-\frac{2}{10}=>\frac{-2}{3}x=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{-2}{3}=>x=\frac{1}{10}.\frac{-3}{2}=>x=\frac{-3}{20}\)
2a)\(\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14}=\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{18}{7}=\frac{1.5}{3.7}-\frac{7.18}{27.7}=\frac{5}{21}-\frac{1.2}{3.1}=\frac{5}{21}-\frac{2}{3}\)
\(\frac{5}{21}-\frac{14}{21}=\frac{-9}{21}=\frac{-3}{7}\)
b) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)
\(0+\left(\frac{15}{26}-\frac{4}{26}\right)=0+\frac{11}{26}=\frac{11}{26}\)