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Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được
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
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
Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\)
\(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)
..........................
\(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)
=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)
=> =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)
Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)
\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)
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=\(\frac{12}{37}:4=\frac{12}{37}.\frac{1}{4}=\frac{3}{37}\)
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
\(A=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)
\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(A=4.\frac{12}{37}\)
\(A=\frac{48}{37}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)
\(=\frac{8}{27}\)
Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{103.107}\)
=\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{103.107}\)
=\(\frac{1}{3.107}\)
=\(\frac{1}{321}\)
k mk nha bn
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