4/11<.../14<5/11
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=\left(-\dfrac{3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
\(=\left(-\dfrac{3}{7}+\dfrac{4}{11}\right).\dfrac{11}{7}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right).\dfrac{11}{7}\)
\(=\dfrac{11}{7}\left[\left(-\dfrac{3}{7}\right)+\dfrac{4}{11}+\left(-\dfrac{4}{7}\right)+\dfrac{7}{11}\right]\)
\(=\dfrac{11}{7}\left[\left(-\dfrac{3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\right]\)
\(=\dfrac{11}{7}\left[\left(-1\right)+1\right]\)
\(=\dfrac{11}{7}.0=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
7/11.3/4+7/11.1/4+4/11
=7/11.(3/4+1/4)+4/11
=7/11.1+4/11
=7/11+4/11
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
=-17/44:7/11+5/77:7/11
=(-1 7/44+5/77):7/11
=-9/28:7/11
=-99/196
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{11}{4}\right)^{12}:\left(-\dfrac{11}{4}\right)^{11}\)
\(=\left(-\dfrac{11}{4}\right)^{12}:\left(-\dfrac{11}{4}\right)^{11}\)
\(=\left(-\dfrac{11}{4}\right)^{12-11}\)
\(=\left(-\dfrac{11}{4}\right)^1\)
\(=-\dfrac{11}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{-15}{4}.\dfrac{9}{11}+\dfrac{15}{4}.\dfrac{15}{11}-\dfrac{15}{4}.\dfrac{6}{11}\)
= \(\dfrac{15}{4}.\dfrac{-9}{11}+\dfrac{15}{4}.\dfrac{15}{11}-\dfrac{15}{4}.\dfrac{6}{11}\)
= \(\dfrac{15}{4}.\left(\dfrac{-9}{11}+\dfrac{15}{11}-\dfrac{6}{11}\right)\)
= \(\dfrac{15}{4}.\left(\dfrac{6}{11}-\dfrac{6}{11}\right)\)
= \(\dfrac{15}{4}.\left(\dfrac{0}{11}\right)\)
= \(\dfrac{15}{4}.0\)
= 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{-3-5-7}{4}=\dfrac{-15}{4}\)
b: \(=\dfrac{-6-5-2+6}{11}=\dfrac{-7}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy: tử của các thừa số trong tích trên là các số nguyên tăng dần bắt đầu từ -4 đến 4 nên sẽ có 1 thừa số = 0/11 = 0
=> tích trên = 0
Lời giải:
$\frac{4}{11}|< \frac{....}{14}< \frac{5}{11}$
$\frac{4\times 14}{11\times 14}< \frac{11\times...}{14\times 11}< \frac{5\times 14}{11\times 14}$
$\frac{56}{11\times 14}< \frac{11\times...}{11\times 14}< \frac{70}{11\times 14}$
$\Rightarrow 56< 11\times....< 70$
$\Rightarrow ... = 6$