\(x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}=30\frac{1}{9}\)

\(x=31\)

\(30\frac{1}{9}\)= \(\frac{271}{9}\)

\(x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}=\frac{271}{9}\)

\(x=\frac{271}{9}-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right)\)

\(x=\frac{263}{9}\)

~ Chúc bạn học tốt ~

X - 2/3 - 2/15 - 2/35 - 2/63 = 30 1/9

X - 2/3 - 2/15 - 2/35 - 2/63 = 271/9

X= 271/9 + 2/63 + 2/35 + 2/15+ 2/3

X = 31

\(\frac{32}{9}\)

2/3+14/15+34/35+62/63=32/9

1-1/2+1/2-1/3+1/3+1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)( chỗ này tối giản nha )

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)
~ Hok tốt ~

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)

\(=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

~ Hok tốt ~

=7/2.(2/1.3+2/3.5+....+2/13.15)

=7/2.(1/1-1/3+1/3-1/5+....+1/13-1/15)

=7/2.(1-1/15)

=7/2.14/15=7.14/2.25

=7.7/1/15=49/15

Ai làm nhanh mk cho nah

\(\frac{7}{1\cdot3}+\frac{7}{3\cdot5}+\frac{7}{5\cdot7}+\frac{7}{7\cdot9}+\frac{7}{9\cdot11}+\frac{7}{11\cdot13}+\frac{7}{13\cdot15}\)

\(=\frac{7}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+..........+\frac{2}{13\cdot15}\right)\)

\(=\frac{7}{2}\left(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+.........+\frac{2}{13}-\frac{2}{15}\right)\)

\(=\frac{7}{2}\left(2-\frac{2}{15}\right)=\frac{98}{15}\)

\(C=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

\(C=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{13}\)

\(C=1-\frac{1}{13}\)

\(C=\frac{12}{13}\)

\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(C=\frac{1}{1}-\frac{1}{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(C=\frac{1}{1}-\frac{1}{13}\)

\(C=\frac{12}{13}\)

\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(C=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{11}-\frac{1}{13}\right)\)

\(C=1-\frac{1}{13}\)

\(C=\frac{12}{13}\)

a) 128 - 3(x + 4) = 23

             3(x + 4) = 128 -23

             3(x + 4) = 105

                x + 4  = 105 : 3

                x + 4  = 35

                x        = 35 - 4

                x        = 31

\(a.128-3\left(x+4\right)=23.\)

\(3\left(x+4\right)=128-23\)

\(3\left(x+4\right)=105\)

\(x+4=105:3=35\)

\(x=35-4=31\)

\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)

\(\left(4x+28\right)\cdot3=175-55=120\)

\(4x+28=120:3=40\)

\(4x=40-28=12\)

\(x=12:4=3\)

a, \(128-3.\left(x+4\right)=23\)

\(\Rightarrow\) \(3.\left(x+4\right)=105\)

\(\Rightarrow\) \(x+4=35\)

\(\Rightarrow\) \(x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(\Rightarrow\) \(\left(4x+28\right).3+55=175\)

\(\Rightarrow\) \(\left(4x+28\right).3=120\)

\(\Rightarrow\) \(4x+28=40\)

\(\Rightarrow\) \(4x=12\)

\(\Rightarrow\) \(x=3\)