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a)MSC: 63. Ta có: \(\frac{-4}{7}\)=\(\frac{-4.9}{7.9}\)=\(\frac{-36}{63}\) ; \(\frac{8}{9}\)=\(\frac{8.7}{9.7}\)=\(\frac{56}{63}\) ; \(\frac{-10}{21}\)=\(\frac{-10.3}{21.3}\)=\(\frac{-30}{63}\)
b) \(\frac{5}{2^2.3}\)=\(\frac{5}{12}\);\(\frac{7}{2^3.11}\)=\(\frac{7}{88}\)
MSC: 264. Ta có: \(\frac{5}{12}\)=\(\frac{5.22}{12.22}\)=\(\frac{110}{264}\) ; \(\frac{7}{88}\)=\(\frac{7.3}{88.3}\)=\(\frac{21}{264}\)
Ta có:
\(\frac{5}{2^2.3}=\frac{5.2.11}{2^2.3.2.11}=\frac{110}{264}\)
\(\frac{7}{2^3.11}=\frac{7.3}{2^3.11.3}=\frac{21}{264}\)
Ta có:
\(\frac{5}{2^{2.}.3}=\frac{5.2.11}{2^2.3.2.11}=\frac{110}{264}\)
\(\frac{7}{2^3.11}=\frac{7.3}{2^3.11.3}=\frac{21}{264}\)
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(\frac{-40}{8}\) và \(\frac{3}{8}\) ; \(\frac{-3}{9}\) và \(\frac{7}{9}\)
1=\(\frac{72}{72}\),-5 = \(\frac{-360}{72}\), \(\frac{3}{8}\)= \(\frac{27}{72}\), \(\frac{-1}{3}\)= \(\frac{-24}{72}\), \(\frac{-7}{-9}\)= \(\frac{56}{72}\)