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b)
\(F=\frac{17^{13}.2^{13}}{34^{12}}=\frac{34^{13}}{34^{12}}=34^1=34.\)
Chúc bạn học tốt!
Bài giải
a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)
\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7 + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15
= 0 -0-0-0-0+7/9 +13/15
= 74/45
a) \(10\sqrt{0,01}.\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac{1}{6}\sqrt{4}\)
\(=10\sqrt{\frac{10}{100}}.\sqrt{\frac{4^2}{3^2}}+3.\sqrt{7^2}-\frac{1}{6}\sqrt{2^2}\)
\(=10.\frac{\sqrt{10}}{10}.\frac{4}{3}+3.7-\frac{1}{6}.2\)
\(=\frac{4\sqrt{10}}{3}+27-\frac{1}{3}\)
\(=\frac{4}{3}\sqrt{10}+\frac{80}{3}\)
b) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(0,8-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
\(=\frac{17}{12}.\frac{1}{400}\)
\(=\frac{17}{4800}\)
e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)
\(=\frac{11}{42}:\frac{1}{28}-8\)
\(=\frac{22}{3}-8\)
\(=-\frac{2}{3}\)
e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)
\(=\frac{11}{42}:\frac{1}{28}-8\)
\(=\frac{22}{3}-8\)
\(=-\frac{2}{3}\)
Bài này mình mới nghĩ ra được đến đây thôi à, các bạn xem rồi giúp mình giải phần tiếp theo nhé !
A = \(\frac{7}{3.4}\) - \(\frac{9}{4.5}\) + \(\frac{11}{5.6}\) - \(\frac{13}{6.7}\) + \(\frac{15}{7.8}\) - \(\frac{17}{8.9}\) + \(\frac{19}{9.10}\)
A = \(\frac{6+1}{3.4}\) - \(\frac{10-1}{4.5}\) + \(\frac{10+1}{5.6}\) - \(\frac{14-1}{6.7}\) + \(\frac{14+1}{7.8}\) - \(\frac{18-1}{8.9}\) + \(\frac{18+1}{9.10}\)
A = \(\frac{6}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{10}{4.5}\) - \(\frac{1}{4.5}\) + \(\frac{10}{5.6}\) + \(\frac{1}{5.6}\) - \(\frac{14}{6.7}\) - \(\frac{1}{6.7}\) + \(\frac{14}{7.8}\) + \(\frac{1}{7.8}\) - \(\frac{18}{8.9}\) - \(\frac{1}{8.9}\) + \(\frac{18}{9.10}\) + \(\frac{1}{9.10}\)
Mình làm được đến đây thôi, các bạn giúp mình nhé ! Thanks !
Đường cùng rồi làm đại thôi, từ đề suy ra nha
=> 7.9.11.13.15.17.19.(\(\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}-\frac{1}{6.7}+\frac{1}{7.8}-\frac{1}{8.9}+\frac{1}{9.10}\) )
chịu thua
a) ta có: \(\frac{x+13}{2006}+\frac{x+2006}{13}+\frac{x+1}{2018}+3=0\)
\(\Rightarrow\frac{x+13}{2006}+1+\frac{x+2006}{13}+1+\frac{x+1}{2018}+1=0\)
\(\Rightarrow\frac{x+2019}{2006}+\frac{x+2019}{13}+\frac{x+2019}{2018}=0\)
\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}\right)=0\)
mà \(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}>0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=-2019\)
b) \(\frac{4}{\left(x+3\right)\left(x+7\right)}+\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{\left(x+7\right)-\left(x+3\right)}{\left(x+3\right)\left(x+7\right)}+\frac{\left(x+10\right)-\left(x+7\right)}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{7}{\left(x+3\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow x=7\)
Hay lắm lên đây hỏi cơ đấy