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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/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
1: B là số nguyên
=>n-3 thuộc {1;-1;5;-5}
=>n thuộc {4;2;8;-2}
3:
a: -72/90=-4/5
b: 25*11/22*35
\(=\dfrac{25}{35}\cdot\dfrac{11}{22}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)
c: \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
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 \(\dfrac{{ - 1}}{8} > \dfrac{{ - 5}}{{24}}\).
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
a)\(\dfrac{-8}{9}< \dfrac{-7}{9}\\ \dfrac{6}{7}< \dfrac{11}{10}\)
a: \(\dfrac{5}{7}=\dfrac{5\cdot11}{7\cdot11}=\dfrac{55}{77}\)
\(\dfrac{9}{11}=\dfrac{9\cdot7}{11\cdot7}=\dfrac{63}{77}\)
b: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot9}{7\cdot9}=\dfrac{54}{63}\)
\(-\dfrac{12}{54}=\dfrac{-2}{9}=\dfrac{-2\cdot7}{9\cdot7}=-\dfrac{14}{63}\)
c: \(\dfrac{-11}{30}=\dfrac{-11\cdot4}{30\cdot4}=\dfrac{-44}{120}\)
\(\dfrac{-17}{-40}=\dfrac{17}{40}=\dfrac{17\cdot3}{40\cdot3}=\dfrac{51}{120}\)
d: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot3}{7\cdot3}=\dfrac{18}{21}\)
\(\dfrac{-12}{36}=\dfrac{-1}{3}=\dfrac{-1\cdot7}{3\cdot7}=\dfrac{-7}{21}\)
a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)
\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)
mà -21>-22
nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)
b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)
\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)
mà -25>-28
nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)
c: \(\dfrac{-8}{7}< -1\)
\(-1< -\dfrac{2}{5}\)
Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)
d: \(-\dfrac{2}{5}< 0\)
\(0< \dfrac{1}{3}\)
Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)