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\(C=\frac{4}{3x7}+\frac{4}{7x11}+\frac{4}{11x15}+\frac{4}{15x19}+\frac{4}{19x23}+\frac{4}{23x27}\)
= 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27
=1/3-(1/7+1/7)-(1/11+1/11)-(1/15-1/15)-(1/19+1/19)-(1/23-1/23)-1/27
=1/3-1/27
=...
=8/27
\(A=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\)(Dấu . là dấu nhân)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}\)
\(=\dfrac{9}{27}-\dfrac{1}{27}\)
\(=\dfrac{8}{27}\)
A = 4/3x7 + 4/7x11+ 4/11x15 + 4/15x19 + 4/19 x23 + 4/23 x 27
A = 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23 -1/27
A = 1/3 - 1/27
A = 8/27
Mình sửa lại đề một chút nhé.
\(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}\)
\(=\dfrac{8}{27}\).
A = \(\dfrac{4}{3\times7}\) + \(\dfrac{4}{7\times11}\) + \(\dfrac{4}{11\times15}\) + \(\dfrac{4}{15\times19}\) + \(\dfrac{4}{19\times23}\) + \(\dfrac{4}{23\times27}\)
A =1/3 -1/7+1/7-1/11 + 1/11-1/15 + 1/15 - 1/19 + 1/19 -1/23+1/23-1/27
A = 1/3 - 1/27
A = 8/27
\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)
a) \(\frac{16}{35}+\frac{8}{35}=\frac{24}{35}\)
b)\(\frac{160}{77}-\frac{28}{77}=\frac{132}{77}=\frac{12}{1}=12\)
c)\(\frac{72}{180}=\frac{18}{45}\)
d) \(\frac{90}{360}=\frac{1}{4}\)
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{12}{15}+\frac{22}{33}+\frac{48}{64}\)
\(=\frac{11}{33}+\frac{16}{64}+\frac{3}{15}+\frac{12}{15}+\frac{22}{33}+\frac{48}{64}\)
\(=\left(\frac{11}{33}+\frac{22}{33}\right)+\left(\frac{16}{64}+\frac{48}{64}\right)+\left(\frac{3}{15}+\frac{12}{15}\right)\)
\(\frac{33}{33}+\frac{64}{64}+\frac{15}{15}\)
\(=1+1+1\)
\(=3\)
Dấu chấm là dấu nhân nhé
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
= \(\frac{1}{3}\)-\(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{11}\)+............+\(\frac{1}{23}\)- \(\frac{1}{27}\)
=\(\frac{1}{3}\)+ 0 + 0 +............+ 0 - \(\frac{1}{27}\)
= \(\frac{1}{3}\)- \(\frac{1}{27}\)
= \(\frac{8}{27}\)