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a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
a)
Ta có: \(BCNN\left( {10,15} \right) = 30\) nên
\(\begin{array}{l}\dfrac{7}{{10}} = \dfrac{{7.3}}{{10.3}} = \dfrac{{21}}{{30}}\\\dfrac{{11}}{{15}} = \dfrac{{11.2}}{{15.2}} = \dfrac{{22}}{{30}}\end{array}\)
Vì \(21 < 22\) nên \(\dfrac{{21}}{{30}} < \dfrac{{22}}{{30}}\) do đó \(\dfrac{7}{{10}} < \dfrac{{11}}{{15}}\).
b)
Ta có: \(BCNN\left( {8,24} \right) = 24\) nên
\(\dfrac{{ - 1}}{8} = \dfrac{{ - 1.3}}{{8.3}} = \dfrac{{ - 3}}{{24}}\)
Vì \( - 3 > - 5\) nên \(\dfrac{{ - 3}}{{24}} > \dfrac{{ - 5}}{{24}}\) do đó \(\dfrac{{ - 1}}{8} > \dfrac{{ - 5}}{{24}}\).
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)
\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)
b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)
a)\(=\dfrac{211}{180}\)
b)\(=\dfrac{5}{39}\)
c)=\(=-\dfrac{65}{168}\)
a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)
nên -11<x<-8
hay \(x\in\left\{-10;-9\right\}\)
b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)
Suy ra: 9<x<14
hay \(x\in\left\{10;11;12;13\right\}\)
c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)
nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)
Suy ra: -536<x<-63
hay \(x\in\left\{-535;-534;...;-64\right\}\)
\(a,-\dfrac{11}{18}< 0;\dfrac{1}{24}>0.Nên:-\dfrac{11}{28}< 0< \dfrac{1}{24}.Vậy:\dfrac{-11}{28}< \dfrac{1}{24}\\ b,\dfrac{-3}{20}< 0;\dfrac{6}{15}>0.Nên:\dfrac{-3}{20}< 0< \dfrac{6}{15}.Vậy:\dfrac{-3}{20}< \dfrac{6}{15}\\ c,\dfrac{-8}{15}=\dfrac{-8.8}{15.8}=\dfrac{-64}{120};\dfrac{-7}{15}=\dfrac{-7.8}{15.8}=\dfrac{-56}{120};\dfrac{x}{40}=\dfrac{3x}{120}\\ Nên:\dfrac{-64}{120}< \dfrac{3x}{120}< \dfrac{-56}{120}.Nên:-64< 3x< -56\\ Nên:-\dfrac{64}{3}< x< -\dfrac{56}{3}\\ \Rightarrow x=19;x=20;x=21\)