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\(\frac{2}{3}:\left(\frac{2}{1}-\frac{2}{3}-\frac{2}{5}-\frac{2}{7}-\frac{2}{9}-\frac{2}{11}\right)x\frac{1}{2}\)
\(=\frac{211}{1155}\)
a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)
\(B=\frac{1}{7}x\left(\frac{34}{11}-\frac{9}{11}-1\right)\)
B = \(\frac{1}{7}x\left(\frac{34-9-11}{11}\right)\)
B = \(\frac{1}{7}x\frac{14}{11}\)
B = \(\frac{2}{11}\)
\(6.x=\frac{1}{2}\)
\(x=\frac{1}{2}:6\)
\(x=\frac{1}{12}\)
\(x:\frac{3}{7}=\frac{7}{11}\)
\(x=\frac{7}{11}.\frac{3}{7}\)
\(x=\frac{3}{11}\)
Dấu chấm là dấu nhân nha
2 x A = 1 - \(\dfrac{1}{2027}\)
\(A=\dfrac{1013}{2027}\)
\(A=\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+...+\dfrac{1}{87\times89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{87}-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{87}-\dfrac{1}{87}\right)-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{89}\)
\(A=\dfrac{84}{445}\)
Vậy, `A=84/445.`
A = \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+\(\dfrac{1}{9\times11}\)+...+\(\dfrac{1}{87\times89}\)
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{2}{5\times7}\)+\(\dfrac{2}{7\times9}\)+\(\dfrac{2}{9\times11}\)+...+\(\dfrac{2}{87\times89}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{87}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{84}{445}\)
A = \(\dfrac{42}{445}\)
\(\frac{1}{7}\times3\frac{1}{11}-\frac{1}{7}\times\frac{9}{11}-\frac{1}{7}\)
\(=\frac{1}{7}\times\frac{34}{11}-\frac{1}{7}\times\frac{9}{11}-\frac{1}{7}\)
\(=\frac{1}{7}\left(\frac{34}{11}-\frac{9}{11}-\frac{11}{11}\right)\)
\(=\frac{1}{7}\times\frac{14}{11}\)
\(=\frac{2}{11}\)
\(\frac{1}{7}\)\(\times3\frac{1}{11}\)\(-\frac{1}{7}\)\(\times\frac{9}{11}\)\(-\frac{1}{7}\)
Đề bài thế này hả bạn?