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\(\frac{9}{4}-y.\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{9}{4}-y.\frac{5}{6}=\frac{5}{6}\)
\(y.\frac{5}{6}=\frac{9}{4}-\frac{5}{6}\)
\(y.\frac{5}{6}=\frac{17}{12}\)
\(y=\frac{17}{12}:\frac{5}{6}\)
\(y=\frac{17}{10}\)
\(\frac{9}{4}\)- y x \(\frac{5}{6}\)= \(\frac{1}{2}\)+ \(\frac{2}{3}\) = \(\frac{9}{4}\)- \(\frac{1}{2}\)- \(\frac{2}{3}\)= y x \(\frac{5}{6}\)
y x \(\frac{5}{6}\)= \(\frac{13}{12}\)
y = \(\frac{13}{12}\)x \(\frac{6}{5}\)= \(\frac{13}{10}\)
y.5 + y : 1/2 + y : 1/3
= y.5 + y.2 + y.3
= y.(5 + 2 + 3)
= y.10
= 1 x 3 + 2 x 3 + 3 x 3 + 4 x 3 + ...+ 9 x 3
= 3 x ( 1 + 2 + 3 + 4 + ...+ 9)
= 3 x 45
= 135
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
A = \(\frac{6x\left(1+8+27\right)}{6x\left(12+96+324\right)}\)= \(\frac{36}{432}\)=\(\frac{1}{12}\)
1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1028=(1+2+3+4+5+6+7+8+9)x3+1028=45x3+1028=135+1028=1163
\(\frac{9}{4}-y.\frac{5}{6}=\frac{1}{2}+\frac{2}{3}\)
\(\frac{9}{4}-y.\frac{5}{6}=\frac{7}{6}\)
\(y.\frac{5}{6}=\frac{9}{4}-\frac{7}{6}\)
\(y.\frac{5}{6}=\frac{13}{12}\)
\(y=\frac{13}{12}:\frac{5}{6}\)
\(y=\frac{13}{10}\)
9 / 4 - y x 5 / 6 = 1 / 2 + 2 / 3
9 / 4 - y x 5 / 6 = 7 / 6
y x 5 / 6 = 13 /12
y = 13 /10