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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 \(\dfrac{{ - 1}}{8} > \dfrac{{ - 5}}{{24}}\).
\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)
Do \(\dfrac{{ - 11}}{8} < 0\) và \(\dfrac{1}{{24}} > 0\) nên \(\dfrac{{ - 11}}{8} < \dfrac{1}{{24}}\)
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
\(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 11^{21}\)
\(\Rightarrow3^{39}< 11^{21}\)
2/
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/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
Ta có:11/12 và 17/18
⇒11x18<12x17 (198<204)
CHÚC BẠN HỌC TỐT!!!!
a: 3/8=36/96
5/12=40/96
b: -2/9=-24/108
-5/12=-45/108
c: -3/16=-63/336
-5/24=-70/336
-21/56=-126/336
Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
\(\dfrac{14}{21};\dfrac{60}{72}\)
\(\dfrac{60}{72}=\dfrac{5}{6}=\dfrac{5\cdot21}{6\cdot21}=\dfrac{105}{126}\)
\(\dfrac{14}{21}=\dfrac{14\cdot6}{21\cdot6}=\dfrac{84}{126}\)
\(\dfrac{84}{123}< \dfrac{105}{126}\Rightarrow\dfrac{14}{21}< \dfrac{60}{72}\)
cho mình nghi lại nha mn \(\dfrac{-3}{5}+\dfrac{28}{5}\cdot\left(\dfrac{43}{56}+\dfrac{5}{24}-\dfrac{21}{63}\right)\)
\(\dfrac{-11}{24}\&\dfrac{-21}{56}\)
\(\dfrac{-21}{56}=\dfrac{-3}{8}=\dfrac{-9}{24}\)
Do \(\dfrac{-11}{24}< \dfrac{-9}{24}\Rightarrow\dfrac{-11}{24}< \dfrac{-21}{56}\)