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\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
(2/5+2/7-2/11):(3/7-3/11+3/5) =2/5+2/7-2/11.7/3-11/3+5/3=2/1+2/1-2/1.1/3-1/3+1/3=2+1/3=7/3 Em đây mới học lớp 6 nên hay xem kĩ lại và tính bang máy tính
a) \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)
\(=\frac{2}{3}+\frac{1}{3}\)
\(=\frac{3}{3}\)
\(=1\)
a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)
\(=\dfrac{9}{16}\)
1. Tính hợp lí :
a) \(\frac{6}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{14}{11}\)
\(=\frac{5}{7}.\left(\frac{6}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}.\frac{-6}{11}=-\frac{30}{77}\)
b) \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)
\(\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}-\frac{1}{3}\right)\)
\(=\frac{1}{3}.\frac{5}{3}=\frac{5}{9}\)
a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)
b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)
a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)
c: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
\(B=\left(\frac{3}{7}+-\frac{3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{5}{9}\right)+\left(\frac{2}{11}+-\frac{2}{11}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)
\(B=0+0+0+0+0+\frac{9}{16}\)
\(B=0+\frac{9}{16}\)
\(B=\frac{9}{16}\)
\(B=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(-\frac{3}{7}+\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(-\frac{2}{11}+\frac{2}{11}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)
\(B=0+0+0+0+0+\frac{9}{16}=\frac{9}{16}\)
a) \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{2}{3}\)
b) \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}=\frac{1}{3}\)
1) Ta có: \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}=\frac{2.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{2}{3}\)
2) Ta có: \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}=\frac{0.125-\frac{1}{5}+\frac{1}{7}}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}=\frac{1}{3}\)
Chuk pạn hok tốt!