\(\frac{a}{b}\)* 4 + \(\frac{1}{6}\)=\(\frac{17}{6}\)
Giải giúp mình nhé
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Tính tất cả ra thì được:
\(=\frac{\frac{973}{60}}{\frac{139}{60}}:\frac{\frac{1255}{221}}{\frac{1506}{221}}+\frac{5858}{5050}\)
\(=\frac{\frac{139}{60}}{\frac{973}{60}}.\frac{\frac{1506}{221}}{\frac{1255}{221}}+\frac{5858}{5050}\)
Tính tử và mẫu dần rồi ra ( phần này dễ mà )
Ta được: ( mình chỉ lấy 2 chữ số phần thập phân thôi )
\(=\frac{1578}{9209}+\frac{5858}{5050}\)
= 133/100
End
a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)
\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)
\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)
\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)
\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)
Vậy x = \(\frac{1}{3}\)
b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)
\(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)
\(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)
\(x.\frac{5}{6}=\frac{-2}{15}\)
\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)
Vậy x = \(\frac{-4}{25}\)
c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)
\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)
\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)
\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)
\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)
Vậy x = \(\frac{-4}{25}\)
Ủng hộ tớ nha m.n
Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)
Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)
Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\times\frac{4}{3}\times...\times\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(\text{ta có:}\frac{6}{a\left(a+7\right)}+1=\frac{\left(a+1\right)\left(a+6\right)}{a\left(a+7\right)}\text{ do đó:}A=\frac{2.7}{1.8}.\frac{3.8}{2.9}.....\frac{101.106}{100.107}\)
\(=\frac{2.3...101.\left(7.8....106\right)}{1....101.\left(8.9.....107\right)}=\frac{7}{107}\)
\(\frac{a}{b}.4+\frac{1}{6}=\frac{17}{6}\)
\(\frac{a}{b}.4\) \(=\frac{17}{6}-\frac{1}{6}\)
\(\frac{a}{b}.4\) \(=\frac{8}{3}\)
\(\frac{a}{b}=\frac{8}{3}:4\)
\(\frac{a}{b}=\frac{2}{3}\)
\(\frac{a}{b}.4=\frac{17}{6}-\frac{1}{6}\)
\(\frac{a}{b}.4=\frac{16}{6}\)
\(\frac{a}{b}=\frac{16}{6}:4\)
\(\frac{a}{b}=\frac{16}{6}.\frac{1}{4}=\frac{16}{6.4}=\frac{2}{3}\)