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A=2/3*5 + 2/5*7 + 2/7*9 + 2/9*11
A=1/3 - 1/5 +1/5 -1/7 + 1/7 - 1/9 + 1/9 - 1/11
A=1/3 - 1/11
A=8/33
Ta có: \(G=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)
2 + 2 + 2 + 2 + 2
1 . 3 3 . 5 5 .7 7 . 9 9 . 11
= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1
3 3 5 5 7 7 9 9 11
= 1 - 1 = 10
11 11
Nhớ **** cho mình đấy nhé!
Số số hạng của A :
( 218 - 2 ) : 2 + 1 = 109
A = 2 + 4 - 6 - 8 + 10 + 12 - 14 - 16 + .... + 210 + 214 - 216 - 218
A = 2 - 6 + 4 - 8 + 10 - 14 + 12 - 16 + ... + 210 - 216 + 214 - 218
Đến đây bạn xem lại đề giùm mình , hiện tại đang ghép nhóm sao cho hiệu mỗi nhóm là -4 ( ghép đôi ) . Mà 109 ko chia hết cho 2
4 số cuối cùng 210 , 214 , 216 , 218 đã đủ cặp và ko hề dư ra số nào
????
a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)
=\(\frac{1}{50}\)
\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)
\(\)
A = \(\frac{1}{3}+\frac{13}{35}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}+\frac{141}{143}\)
\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(=6-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=6-\left(1-\frac{1}{13}\right)\)
\(=6-1+\frac{1}{13}\)
\(=5+\frac{1}{13}\)
\(=\frac{66}{13}\)
\(\text{Vậy }A=\frac{66}{13}\)
\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)
\(\Rightarrow3x-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)=0\)
\(\Rightarrow3x-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)=0\)
\(\Rightarrow3x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)
\(\Rightarrow3x-\left(1-\frac{1}{99}\right)=0\)
\(\Rightarrow3x-\frac{98}{99}=0\)
\(\Rightarrow3x=0+\frac{98}{99}\)
\(\Rightarrow3x=\frac{98}{99}\)
\(\Rightarrow x=\frac{98}{99}:3\)
\(\Rightarrow x=\frac{98}{297}\)
\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)
\(2\left(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}\right)=2.0\)
\(6x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}=0\)
\(6x-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=0\)
\(6x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)
\(6x-\left(1-\frac{1}{11}\right)=0\)
\(6x-\frac{10}{11}=0\)
\(6x=\frac{10}{11}\)
\(x=\frac{5}{33}\)
A=\(\frac{\left(1+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
A=\(\frac{\left(1+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).0}{1-2+3-4+...+99-100}\)
A= 0
KẾT QUẢ ĐÚNG 100%
\(\frac{2}{7}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)