Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)
= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))
= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))
= 12 x \(\frac{1}{2}\) x ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
= 6 x ( \(\frac{1}{3}\) - \(\frac{1}{11}\))
= 6 x \(\frac{8}{33}\)
= \(\frac{48}{33}\)=\(\frac{16}{11}\)
Nhớ tk nha
a) \(\frac{1212}{1515}\):\(\frac{2727}{2525}\)
= \(\frac{1212}{1515}\)* \(\frac{2525}{2727}\)
= \(\frac{101.12}{101.15}\)* \(\frac{101.25}{101.27}\)
= \(\frac{12}{15}\). \(\frac{25}{27}\)
= \(\frac{20}{27}\)
b) ban co the viet ro hon de bai dc ko?
a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)
b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)
\(=\frac{101.10+101.11+...+101.17}{101.20+101.21+...+101.27}\)
\(=\frac{101.\left(10+11+...+17\right)}{101.\left(20+21+...+27\right)}\)
\(=\frac{108}{188}\)
\(=\frac{27}{47}\)
\(2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right)\cdot5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{20}{120}+\frac{16}{120}+\frac{9}{120}+\frac{5}{120}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{5}{12}:5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{1}{12}.y>\frac{5}{6}\)
Đặt :\(\frac{1}{12}.y=2\Rightarrow y=2:\frac{1}{12}=24\)
\(\frac{1}{12}.y=\frac{5}{6}\Rightarrow y=\frac{5}{6}:\frac{1}{12}=10\)
\(\Rightarrow24>y>10\)
\(\Rightarrow y\in\left\{11;12;...;23\right\}\)
Gợi ý: rút gọn cho 101 rồi đặt 5 ra ngoài làm thừa số chung thì sẽ tìm ra kết quả là \(\frac{25}{24}\)
\(\frac{1212}{3434}+x-\frac{13}{102}=1-\frac{107}{102}\)
\(\frac{36}{102}+x-\frac{13}{102}=-\frac{5}{102}\)
\(\frac{23}{102}+x=-\frac{5}{102}\)
\(x=-\frac{5}{102}-\frac{23}{102}=-\frac{14}{51}\)
lại nữa
làm đúng mà dis hoài
bực ơi là bực
bực ko chịu nỗi luôn 😡😡😡
thằng mô dis tui khai mau tui dis hết ko chừa 1 phát nào 😡😡😡
\(\frac{1339}{1442}=\frac{13}{14}\)
\(\frac{1212}{3030}=\frac{2}{5}\)
\(\frac{1339}{1442}=\frac{1300+39}{1400+42}=\frac{13\times100+13\times3}{14\times100+14\times3}=\frac{13\times\left(100+3\right)}{14\times\left(100+3\right)}=\frac{13}{14}\)\(\frac{13}{14}\)
\(\frac{1212}{3030}=\frac{1200+12}{3000+30}=\frac{12\times100+12}{30\times100+30}=\frac{12\times\left(100+1\right)}{30\times\left(100+1\right)}=\frac{12}{30}=\frac{2}{5}\)
\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)
=\(\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)
=\(\dfrac{16}{11}\)
Giải:
\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)
\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)
\(=12.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)\)
\(=12.\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\)
\(=6.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=6.\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\)
\(=6.\dfrac{8}{33}\)
\(=\dfrac{16}{11}\)