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a) \(\frac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\frac{2^{23}}{2^{24}}=\frac{1}{2}\)
Các phần kia tương tự, à bạn đăng 1 2 câu hỏi 1 lần thôi, đăng nhiều quá ko ai trả lời đâu
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Thế à ! Vậy bạn hãy nhấp vào https://h.vn/hoi-dap/question/646555.html?pos=1792187 mà xem
Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)
=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)
\(=\frac{5}{2}.\frac{6}{5}-17\)
\(=3-17=-14\)
Tụi quá mới lớp 5 thui
d) \(\dfrac{8^4.3^6}{2^7.65}=\dfrac{\left(2^3\right)^4.3^6}{2^7.65}=\dfrac{2^{12}.3^6}{2^7.65}=\dfrac{2^7.2^5.3^6}{2^7.65}=\dfrac{2^5.3^6}{65}=\dfrac{23328}{65}\)
c) \(\left(\dfrac{3}{5}-\dfrac{3}{4}\right).\left(\dfrac{2}{6}-\dfrac{1}{5}\right)^2=\dfrac{3.4-3.5}{4.5}.\left(\dfrac{2.5-1.6}{6.5}\right)^2\\ =\dfrac{-3}{20}.\left(\dfrac{2}{15}\right)^2=\dfrac{-3}{20}.\dfrac{4}{225}\\ =\dfrac{-3}{4.5}.\dfrac{4}{75.3}=\dfrac{-1}{375}\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
Bài làm :
\(\text{a)}=2,5-1,65.\frac{10}{11}=2,5-1.5=1\)
\(b\text{)}=\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{2015-2012}{2012.2015}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{2012}-\frac{1}{2015}\)
\(=\frac{1}{5}-\frac{1}{2015}\)
\(=\frac{402}{2015}\)
\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)
\(=\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)
\(=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)