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Z
22 tháng 7 2019
Lời giải:
Ta có: F = \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}+0,2}{\frac{3}{4}+0,5+\frac{3}{10}}\) = \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3.0,125-3.\frac{1}{5}+3.\frac{1}{7}}+\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}{\frac{3}{2}.\frac{1}{2}+\frac{3}{2}.\frac{1}{3}+\frac{3}{2}.\frac{1}{5}}\)
= \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3.\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}{\frac{3}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)}\)
= \(\frac{1}{3}+\frac{1}{\frac{3}{2}}\)
= \(\frac{1}{3}+\frac{2}{3}\)
= \(\frac{3}{3}\) = 1. Vậy: F = 1
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IM
11 tháng 9 2016
\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}+\frac{3}{10}}\)
\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{2}.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}\)
\(\Rightarrow A=\frac{1}{3}+\frac{2}{3}\)
\(\Rightarrow A=1\)
DC
23 tháng 8 2018
=1/8-1/5+1/7*3/8-3/5+3/7 + 1/2+1/3-1/5*3/4+1/2-3/10
=19/280*57/280+19/30*19/20
=19/280.280/57+19/30.20/19
=1/1.1/3+1/3.2/1
=1/3+2/3=3/3
=1
DT
28 tháng 10 2019
\(\frac{1\left(0,125-\frac{1}{5} +\frac{1}{7}\right)}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{1\left(\frac{1}{2}+\frac{1}{3}-0,2\right)}{\frac{3}{2}\left(\frac{1}{2}+\frac{1}{3}-0,2\right)}\)\(=\frac{1}{3}+\frac{1}{\frac{3}{2}}=\frac{1}{3}+\frac{2}{3}=\frac{3}{3}=1\)
Vậy giá trị biểu thức trên bằng 1
VM
28 tháng 10 2019
\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
\(=\frac{\frac{19}{280}}{\frac{57}{280}}+\frac{\frac{19}{30}}{\frac{19}{20}}\)
\(=\frac{1}{3}+\frac{2}{3}\)
\(=1.\)
Chúc bạn học tốt!
26 tháng 3 2017
(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
28 tháng 5 2022
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
17 tháng 7 2019
a) \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)
\(=\frac{2}{3}+\frac{1}{3}\)
\(=\frac{3}{3}\)
\(=1\)
PQ
3
27 tháng 6 2016
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) \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}=\frac{1}{3}\)
NM
27 tháng 6 2016
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! 