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2) = -1/2 . -2/3 .-3/4 ..... . -98/99 = 1/99 (Tích này có 98 thừa số âm, 98 là số chẵn nên tích mang dấu dương)
\(A=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{21}+\frac{9}{53} \)
\(A=\left(\frac{16}{7}+\frac{10}{21}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)+\frac{21}{31}\)
\(A=\frac{58}{21}+1+\frac{21}{31}\)
\(A=\frac{100}{21}\)
\(B=6\left(\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=6.\frac{1}{18}\)
\(B=\frac{1}{3}\)
Ta có: \(A=\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\)
\(=\frac{1}{3}(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90})\)
\(=\frac{1}{3}(\frac{1}{15}-\frac{1}{90})\)
\(=\frac{1}{3}(\frac{6}{90}-\frac{1}{90})\)
\(=\frac{1}{3}.\frac{5}{90}\)
\(=\frac{1}{54}\)
Ta có: 1= \(\frac{54}{54}\)
Suy ra A < 1 (đpcm)
3A=3*(1/15*18+1/18*21+...+1/87*90)
3A=3/15*18+3/18*21+...+3/87*90
3A=1/15-1/18+1/18-1/21+...+1/87-1/90
3A=1/15-1/90
3A=1/18
A=1/18 chia3
A=1/54
vì 1/54<1 nên A<1
Ta có A = \(\frac{4}{3.7}+\frac{4}{7.11}+..............+\frac{4}{107.111}\)
=> A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.............+\frac{1}{107}-\frac{1}{111}\)
A = \(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)
k nha bạn
\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(\Rightarrow=\frac{1}{3}-\frac{1}{111}\)
\(=\frac{12}{37}\)
k nha
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(=\frac{1}{3}-\frac{1}{111}\)
\(=\frac{108}{333}=\frac{12}{37}\)
\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)
\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(A=4.\frac{12}{37}\)
\(A=\frac{48}{37}\)
Giải:
a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\frac{1}{18}\)
C = \(2.\frac{1}{18}\)
C = \(\frac{1}{9}\)
Vậy C = \(\frac{1}{9}\)
b) D = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
D = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)\
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\frac{2}{75}\)
D = \(\frac{1}{75}\)
Vậy D = \(\frac{1}{75}\)
c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)
E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)
E = \(\frac{1}{8}-\frac{1}{41}\)
E = \(\frac{33}{328}\)
Vậy E = \(\frac{33}{328}\)
câu 1
\(\Leftrightarrow A=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+...+\frac{4}{107}-\frac{4}{111}\)
\(\Rightarrow A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(\Rightarrow A=4.\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(\Rightarrow A=4.\frac{12}{37}\)
\(\Rightarrow A=\frac{48}{37}\)
phần B làm tương tự
câu 2:
a)\(\Leftrightarrow x+\frac{7}{12}=\frac{15}{18}\)
\(\Rightarrow x=\frac{15}{18}-\frac{7}{12}\)
\(\Rightarrow x=\frac{1}{4}\)
b,c tương tự như câu 1 phần a
Câu 1:
Ta có: A=1/3-1/7+1/7-1/11+....+1/107-1/111
=> A=1/3+(-1/7+1/7)+(-1/11+1/11)+....+(-1/107+1/107)+(-1)/111
=>A=1/3+(-1)/111
=>A=12/37
Ta có B= 6(1/15.18+1/18.21+...+1/87.90)
=> 3B= 6(3/15.18+3/18.21+...+3/87.90)
=> 3B= 6(1/15-1/18+1/18-1/21+....+1/87-1/90)
(Tương tự như câu A) 3B=6[1/15+(-1)/90]
=> 3B= 6.1/18=1/3
=> B= 1/3:3 = 1/9