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\(M=\dfrac{26}{7.11}+\dfrac{65}{11.21}+\dfrac{39}{21.27}+\dfrac{143}{27.49}\)
\(M=\dfrac{13.2}{7.11}+\dfrac{13.5}{11.21}+\dfrac{13.3}{21.27}+\dfrac{13.11}{27.49}\)
\(M=13.\dfrac{2}{7.11}+13.\dfrac{5}{11.21}+13.\dfrac{3}{21.27}+13.\dfrac{11}{27.49}\)
\(M=13.\left(\dfrac{2}{7.11}+\dfrac{5}{11.21}+\dfrac{3}{21.27}+\dfrac{11}{27.49}\right)\)
\(M=13.\dfrac{1}{2}\left(\dfrac{4}{7.11}+\dfrac{10}{11.21}+\dfrac{6}{21.27}+\dfrac{22}{27.49}\right)\)
\(M=\dfrac{13}{2}\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\)
\(M=\dfrac{13}{2}.\dfrac{6}{49}=\dfrac{39}{49}\)
N sai đề rồi chuyển 35 thành 85 tính tương tự xong lấy \(P=\dfrac{M}{N}+\dfrac{34}{29}\)
P=2
M=26/7x11+65/11x21+39/21x27+143/37x49
M=26/77+65/231+13/189+143/1813
M=0,766705481
N=51/7x16+17/16x19+35/19x34+85/34x49
N=51/112+17/304+35/646+5/98
N=0,6164781702
P=0,766705481/0,6164781702+29/34
P=2,096627519
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(=13\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
\(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+......-\frac{1}{49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right);B=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-.....-\frac{1}{49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\Rightarrow\frac{A}{B}=\frac{17}{13}\)
Có: \(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Ttự, ta đc: \(B=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Vậy \(\frac{A}{B}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
#Walker
