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4A=\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)
4A=\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
4A=\(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)
A=\(\frac{12}{37}:4=\frac{12}{37}.\frac{1}{4}=\frac{3}{37}\)
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
\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)
\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(A=4.\frac{12}{37}\)
\(A=\frac{48}{37}\)
4x(\(\frac{1}{3.7}+...+\frac{1}{107.111}\) )
4(\(\frac{1}{3}-\frac{1}{7}+...+\frac{1}{107}-\frac{1}{111}\))
4(\(\frac{1}{3}-\frac{1}{111}\))
4.\(\frac{12}{37}\)
48/37
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}\)
\(=\frac{12}{25}\)
\(B=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{23.27}\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{4}.\frac{8}{27}=\frac{2}{27}\)
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
A=1/3*7+1/7*11+..+1/95*99
=> 4A=4/3*7+4/7*11+..+4/95*99
=>4A=1/3-1/7+1/7-1/11+...+1/95-1/99=1/3-1/99=32/99
=>A=8/99
\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.......+\frac{4}{95.99}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)
a) 4/ 3x7 + 4/7x11+ 4/11x15+...+ 4/107x111
=1/3-1/7+ 1/7-1/11+ 1/11- 1/15+...+1/107 - 1/111
= 1/3-1/111
=12/37
\(b,\frac{3^2}{8\cdot11}+\frac{3^2}{11\cdot14}+\frac{3^2}{14\cdot17}+...+\frac{3^2}{197\cdot200}\)
\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\cdot\frac{3}{25}=\frac{9}{25}\)