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\(\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+...+\frac{1}{99.100}\)
= \(2.\left(\frac{1}{1.3.2}+\frac{1}{3.2.2}+\frac{1}{2.5.2}+...+\frac{1}{99.50.2}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{100}\right)\)
= \(2.\frac{49}{100}\)
= \(\frac{49}{50}\)
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
1, =35 x 77-35 x 12+35x35-35 x 18 suy ra =35 x(77-12+35-18)=35 x 82=2870.
2, =2014 x (100-1)=201400-2014=199386.
3,=35x12+65x12+65=12x(65+35)+65=1200+65=1265.
4.=\(\frac{13\times\left(37-1\right)}{12\times\left(37+2\right)}\)=\(\frac{13\times36}{12\times39}=\frac{13\times3\times12}{12\times3\times13}=1\)
tck nha bạn
\(A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{93.96}+\frac{3}{96.99}\)
\(A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{93}-\frac{1}{96}+\frac{1}{96}-\frac{1}{99}\)
\(A=1-\frac{1}{99}=\frac{98}{99}\)
Vậy A=\(\frac{98}{99}\)
\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
\(3B=\)\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(3B=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)
\(B=\frac{24}{49}:3=\frac{8}{49}\)
Vậy B=\(\frac{8}{49}\)
Dấu "." là dấu nhân.
_Học tốt_
\(2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}=1-\frac{1}{15}=\frac{14}{15}\)
\(=2.2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=4.\left(1-\frac{1}{15}\right)\)
\(=4.\frac{14}{15}\)
\(=\frac{56}{15}\)
A = 3/1.3 + 3/3.5 + 3/5.7 + 3/7.9 + ... + 3/97.99
A = 3/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + .... + 2/97 - 2/99
A = 3/2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 )
A = 3/2 . ( 1 - 1/99 )
A = 3/2 . 98/99
A = 49/33
b) dãy số không có quy luật==> bạn xem lại đề
c) \(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{50\times51}-\frac{1}{51\times52}\right)\)
\(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{51\times52}\right)=\frac{1}{2}\times\frac{2650}{5408}=\frac{1325}{5408}\)
a, 1/2x5 - 1/5x8 - 1/8x11 - 1/11x14
= 1/2 x 1/5 - 1/5 x 1/8 - 1/8 x 1/11 - 1/11 x 1/14
= 1/2 - 1/14 = 3/7
b và c mik nghĩ là lỗi đề (vì mik đã được học r). Theo mik nghĩ đề phải như thế này:
b) 1/2 + 1/6 + 1/12 + ... + 1/56 + 1/72
c) 9/10 + 39/40 + 87/88 + 153/154
Bạn làm được chưa?