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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 \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
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) \(\frac{4}{\left(x+3\right)\left(x+7\right)}+\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{\left(x+7\right)-\left(x+3\right)}{\left(x+3\right)\left(x+7\right)}+\frac{\left(x+10\right)-\left(x+7\right)}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow\frac{7}{\left(x+3\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)
\(\Rightarrow x=7\)
1 \(=\)\(\frac{46656}{216}\)\(=\)216
2\(=\)\(\frac{64}{1024}\)\(=\)\(\frac{1}{16}\)
3 \(=\)\(\frac{900}{-27000}\)\(=\)\(\frac{-1}{30}\)
4 \(=\)\(\frac{225}{-3375}\)\(=\)\(\frac{-1}{15}\)
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
\(\frac{y+9}{35}+\frac{y+11}{34}=\frac{y+13}{33}+\frac{y+15}{32}\)
\(\Leftrightarrow\frac{y+9}{35}+2+\frac{y+11}{34}+2=\frac{y+13}{33}+2+\frac{y+15}{32}+2\)
\(\Leftrightarrow\frac{y+79}{35}+\frac{y+79}{34}=\frac{y+79}{33}+\frac{y+79}{32}\)
\(\Leftrightarrow\frac{y+79}{35}+\frac{y+79}{34}-\frac{y+79}{33}-\frac{y+79}{32}=0\)
\(\Leftrightarrow\left(y+79\right)\left(\frac{1}{35}+\frac{1}{34}-\frac{1}{33}-\frac{1}{32}\right)=0\)
\(\Leftrightarrow y+79=0\).Do \(\frac{1}{35}+\frac{1}{34}-\frac{1}{33}-\frac{1}{32}\ne0\)
\(\Leftrightarrow y=-79\)
Mk làm bài này trên cơ sở bài làm của bạn:
\(A=\frac{7}{3.7}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{10}{9.10}\)
\(A=\frac{8-1}{3.4}-\frac{10-1}{4.5}+\frac{12-1}{5.6}-\frac{14-1}{6.7}+\frac{16-1}{7.8}-\frac{18-1}{8.9}+\frac{20-1}{9.10}\)
\(A=\frac{8}{3.4}-\frac{1}{3.4}+\frac{12}{5.6}-\frac{1}{5.6}+\frac{14}{6.7}-\frac{1}{6.7}+\frac{16}{7.8}-\frac{1}{7.8}+\frac{18}{8.9}-\frac{1}{8.9}+\frac{20}{9.10}-\frac{1}{9.10}\)
\(A=\left(\frac{8}{3.4}+\frac{12}{5.6}+\frac{14}{6.7}+\frac{16}{7.8}+\frac{18}{8.9}+\frac{20}{9.10}\right)-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\left(\frac{2.4}{3.4}+\frac{2.6}{5.6}+\frac{2.7}{6.7}+\frac{2.9}{8.9}+\frac{2.10}{9.10}\right)-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(\frac{2}{3}+\frac{2}{5}+\frac{2}{6}+\frac{2}{8}+\frac{2}{9}\right)-\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(A=\left[\left(\frac{2}{3}+\frac{2}{9}\right)+\left(\frac{2}{6}+\frac{2}{8}\right)+\frac{2}{5}\right]-\frac{7}{30}\)
\(A=\left(\frac{8}{9}+\frac{7}{12}+\frac{2}{5}\right)-\frac{7}{30}\)
\(A=\left(\frac{160}{180}+\frac{105}{180}+\frac{72}{180}\right)-\frac{42}{180}\)
\(A=\frac{337}{180}-\frac{42}{180}\)
\(A=\frac{295}{180}=\frac{59}{36}\)
Nguyễn Huy Tú : Cô mình nói đáp án đúng là \(\frac{13}{30}\) còn đáp án của bạn khác với đáp án của cô mình.
b)
\(F=\frac{17^{13}.2^{13}}{34^{12}}=\frac{34^{13}}{34^{12}}=34^1=34.\)
Chúc bạn học tốt!