\(A=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

\(A=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(A=5.\left(1-\frac{1}{31}\right)\)

\(A=5.\frac{30}{31}\)

\(A=\frac{150}{31}>1\)

Đề hơi lạ nhỉ, vì quá rõ ràng rùi 5²/1.6 = 25/6 > 1 nên A lớn hơn 1

\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

   \(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

   \(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

   \(=5\left(1-\frac{1}{31}\right)\)

    \(=5.\frac{30}{31}=\frac{150}{31}>1\)

=> A > 1

Study well ! >_<

#)Giải :

Ta có :

\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)

\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)=5\left(1-\frac{1}{31}\right)=5\times\frac{30}{31}=\frac{150}{31}>1\)

\(\Rightarrow A>1\)

Ta có: \(A=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{26\cdot31}\)

\(=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{26\cdot31}\right)\)

\(=5\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5\cdot\left(1-\frac{1}{31}\right)=5\cdot\frac{30}{31}=\frac{150}{31}>1\)

hay A>1(đpcm)

Q=5(5/1x6+5/6x11+5/11x16+....+5/26x31)

Q=5(1/1-1/6+1/6-1/11+1/11-1/16+....+1/26-1/31)

Q=5(1/1-1/31)

Q=5x30/31

Q=150/31

\(Q=\frac{25}{1.6}+\frac{25}{6.11}+\frac{25}{11.16}+......+\frac{25}{26.31}.\)

\(Q=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{26}-\frac{1}{31}\right)\)

\(Q=5\left(1-\frac{1}{31}\right)\)

CÒN ĐÔU PN TỰ LÀM NHA

\(Q=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)

\(Q=5\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{26.31}\right)\)

\(Q=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(Q=5\left(1-\frac{1}{31}\right)\)

\(Q=5.\frac{31}{32}\)

\(Q=\frac{155}{32}\)

\(A=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(A=\frac{5}{1}-\frac{5}{6}+\frac{5}{6}-\frac{5}{11}+...+\frac{5}{26}-\frac{5}{31}\)
\(A=5-\frac{5}{31}=\frac{150}{31}>1\left(150>31\right)\)

\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)

\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)

\(a\)) Mình giải theo cách khác:

Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)

Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

\(S=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

\(S=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(S=5.\left(1-\frac{1}{31}\right)\)

\(S=5.\frac{30}{31}\)

\(S=\frac{150}{31}\)

Câu L bạn thiếu số\(\frac{1}{475}\)

\(L=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

\(L=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)

\(L=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)

\(L=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)

\(L=\frac{1}{6}.\left(1-\frac{1}{37}\right)\)

\(L=\frac{1}{6}.\frac{36}{37}\)

\(L=\frac{6}{37}\)

\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

\(=5\left[\left(1-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+...+\left(\frac{1}{26}-\frac{1}{31}\right)\right]\)

\(=5\left[1-\frac{1}{31}\right]\)

\(=5.\frac{30}{31}=\frac{150}{31}\)

\(L=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{775}+\frac{1}{1147}\)

\(L.6=\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}\)

\(L.6=\left(1-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)+\left(\frac{1}{19}-\frac{1}{25}\right)+\left(\frac{1}{25}-\frac{1}{31}\right)\)

\(L.6=1-\frac{1}{31}\)

\(L.6=\frac{31}{31}-\frac{1}{31}\)

\(L.6=\frac{25}{31}\)

\(L=\frac{30}{31}:6\)

\(L=\frac{30}{31}.\frac{1}{6}\)

\(L=\frac{30}{186}\)

S=5*5/1*6 + 5*5/6*11+...+5*5/26*31

S=5*(5/1*6+5/6*11+...+5/26*31)

S=5*(1-1/6+1/6-1/11+...+1/26-1/31)

S=5*(1-1/31)

S=5*30/31

S=150/31

Vậy S=150/31

Ta có:

\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)

\(A=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

\(A=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(A=5\left(\frac{1}{1}-\frac{1}{31}\right)\)

\(A=5.\frac{30}{31}\)

\(A=\frac{150}{31}\)

Vậy \(A=\frac{150}{31}\)