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\(\frac{5}{7}>\frac{X}{10}>\frac{4}{7}\)=>\(\frac{50}{70}>\frac{7X}{70}>\frac{40}{70}\)
=> 50>7X>40
=> \(\frac{50}{7}>X>\frac{40}{7}\)
Câu 2.
=>x +4<11
=>x=0;1;2;3;4;5;6
=>x +4>9
=>x=6;7;8;9;...
=> x=6
\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
<=> \(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)
<=> \(94< x< 92\)vô lí
Vậy không tìm đc x thỏa mãn
\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(=\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow94< x< 92\)
\(\Rightarrow\)ĐỀ SAI
a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
(\(\frac{5}{7}-y\)) x \(\frac{14}{5}=\frac{6}{5}\)
\(\frac{5}{7}-y\) = \(\frac{6}{5}:\frac{14}{5}\)
\(\frac{5}{7}-y\) = \(\frac{3}{7}\)
y = \(\frac{5}{7}-\frac{3}{7}\)
y = \(\frac{2}{7}\)
\(\left(\frac{5}{7}-y\right)\cdot\frac{14}{5}=\frac{7}{10}+\frac{1}{2}\)
\(\Leftrightarrow\left(\frac{5}{7}-y\right)\cdot\frac{14}{5}=\frac{7}{10}+\frac{5}{10}\)
\(\Leftrightarrow\left(\frac{5}{7}-y\right)\cdot\frac{14}{5}=\frac{6}{5}\)
\(\Leftrightarrow\frac{5}{7}-y=\frac{6}{5}\div\frac{14}{5}\)
\(\Leftrightarrow\frac{5}{7}-y=\frac{3}{7}\)
\(\Rightarrow y=\frac{5}{7}-\frac{3}{7}=\frac{2}{7}\)
\(\frac{2}{3}\div\frac{x}{5}+\frac{5}{7}=\frac{2}{7}\div\frac{3}{5}+\frac{10}{9}\)
\(\frac{2}{3}\div\frac{x}{5}+\frac{5}{7}=\frac{10}{21}+\frac{10}{9}\)
\(\frac{2}{3}\div\frac{x}{5}+\frac{5}{7}=\frac{300}{189}\)
\(\frac{2}{3}\div\frac{x}{5}=\frac{300}{189}-\frac{5}{7}\)
\(\frac{2}{3}\div\frac{x}{5}=\frac{55}{63}\)
\(\frac{x}{5}=\frac{2}{3}\div\frac{55}{63}\)
\(\frac{x}{5}=\frac{42}{55}\)
\(x=42\div\left(55\div5\right)=\frac{42}{11}\)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=2\)
b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)
c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)
\(=2\div\frac{1}{2}=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}=\frac{3}{2}\)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=1+1\)
\(=2\)
b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)
\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)
\(=\frac{25}{81}.\frac{3}{4}\)
\(=\frac{25}{108}\)
c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)
\(=\frac{7}{8}.2+\frac{9}{8}.2\)
\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)
\(=2.\frac{16}{8}\)
\(=2.2\)
\(=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
=>\(\frac{50}{70}>\frac{x.7}{70}>\frac{40}{70}\)
=>x.7\(\in\) {49;48;47;46;45;44;43;42;41}
49;42\(⋮\) x.7
x.7=42
x=42/7=6
x.7=49
x=49/7=7
tk cho minh ha