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\(\frac{2^2}{15}+\frac{2^2}{35}+\frac{2^2}{63}+\frac{2^2}{99}+\frac{2^2}{143}=2\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\right)=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)=2\cdot\left(\frac{1}{3}-\frac{1}{13}\right)=2\cdot\frac{10}{39}=\frac{20}{39}\)
\(=2\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=2\left(1-\frac{1}{13}\right)=2.\frac{12}{13}=\frac{24}{13}\)
Theo đề ta có:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)
=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)
= -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)
= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)
= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)
= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)
= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)
= \(-\frac{38}{15}+\frac{1}{23}\)
= \(-\frac{859}{345}\)
= \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
= \(\frac{2}{7}.2\)
= \(\frac{4}{7}\)
\(\frac{2}{7}.5\frac{1}{4}-\frac{2}{7}.3\frac{1}{4}\)
=> \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
=> \(\frac{2}{7}.2\)
=> \(\frac{4}{7}\)
#Hk_tốt
#Ngọc's_Ken'z
B=2/15+2/35+2/63+2/99
B=2(1/3.5+1/5.7+1/7.9+1/9.11) khoảng cách từ 3-5;5-5;7-9;9-11 là 2 nen
B=2/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11) gop -1/5+1/5;-1/7+1/7;-1/9+1/9=0
B=1(1/3-1/11)=8/33
=2/3*5+2/5*7+2/7*9+2/9*11
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11
=1/3-1/11
=8/33
= [ 60/90 - ( 12/15 + 10/15) ] : 6/5
= ( 2/3 - 22/15 ) x 5/6
= ( 10/15 - 22/15 ) x 5/6
= -12/15 x 5/6
= -60/90
= -2/3
( 15/10 x 4/9 - ( 4/5 + 2/3 )) : 6/5
( 15/10 x 4/9 - ( 12/15 + 10/15 )) : 6/5
(30/45 - 66/45 ) : 6/5
-12/15 : 6/5 ( đã rút gọn -36/45 = -12/15 )
-2/3
k mk na <3
a. \(2\cdot125\cdot\left(-5\right)\cdot8\)
\(=\left(2\cdot\left(-5\right)\right)\cdot\left(125\cdot8\right)\)
\(=-10\cdot1000\)
\(=-10000\)
b. \(56\cdot\left(-35\right)+156\cdot35\)
\(=-1960+5460\)
\(=3500\)
c. \(\frac{11}{15}+\frac{-9}{10}=\frac{22}{30}+\frac{-27}{30}=\frac{-5}{30}=\frac{-1}{6}\)
d. \(\frac{-4}{5}+\frac{2}{7}+\frac{3}{31}+\frac{5}{7}+\frac{-1}{5}\)
\(=\left(\frac{-4}{5}+\frac{-1}{5}\right)+\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{3}{31}\)
\(=-1+1+\frac{3}{31}\)
\(=0+\frac{3}{31}\)
\(=\frac{3}{31}\)
a, 6/7+5/9+8/7-2/9=(6/7+8/7)+(5/9-2/9)=14/7+3/9=2+1/3=7/3
b, 5+7/3=22/3
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{899}\)
\(=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{29\cdot31}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{31}\)
\(=\frac{1}{3}-\frac{1}{31}\)
\(=\frac{28}{93}\)
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{899}\)
= \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{29.31}\)
= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)
= \(\frac{1}{3}-\frac{1}{31}+0+0+...+0\)
= \(\frac{29}{93}\)