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\(A=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(A=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5.\left(1-\frac{1}{31}\right)\)
\(A=5.\frac{30}{31}\)
\(A=\frac{150}{31}>1\)
Đề hơi lạ nhỉ, vì quá rõ ràng rùi 52/1.6 = 25/6 > 1 nên A lớn hơn 1
Ta có :
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(A=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(A=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5\left(1-\frac{1}{31}\right)\)
\(A=5.\frac{30}{31}\)
\(A=\frac{150}{31}>1\)
\(\Rightarrow\)\(A>1\)
Vậy \(A>1\)
Chúc bạn học tốt ~
Ko cần dài dòng vậy đâu,A=\(\frac{5^2}{1.6}+\left(\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\right)\)
Ta thấy \(\frac{5^2}{1.6}>1\)và tổng trong ngoặc >0 nên =>A>1
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
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(S=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(S=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(S=5.\left(1-\frac{1}{31}\right)\)
\(S=5.\frac{30}{31}\)
\(S=\frac{150}{31}\)
Câu L bạn thiếu số\(\frac{1}{475}\)
\(L=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(L=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(L=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)
\(L=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
\(L=\frac{1}{6}.\left(1-\frac{1}{37}\right)\)
\(L=\frac{1}{6}.\frac{36}{37}\)
\(L=\frac{6}{37}\)
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5\left[\left(1-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+...+\left(\frac{1}{26}-\frac{1}{31}\right)\right]\)
\(=5\left[1-\frac{1}{31}\right]\)
\(=5.\frac{30}{31}=\frac{150}{31}\)
\(L=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{775}+\frac{1}{1147}\)
\(L.6=\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}\)
\(L.6=\left(1-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)+\left(\frac{1}{19}-\frac{1}{25}\right)+\left(\frac{1}{25}-\frac{1}{31}\right)\)
\(L.6=1-\frac{1}{31}\)
\(L.6=\frac{31}{31}-\frac{1}{31}\)
\(L.6=\frac{25}{31}\)
\(L=\frac{30}{31}:6\)
\(L=\frac{30}{31}.\frac{1}{6}\)
\(L=\frac{30}{186}\)
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
\(a\)) 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: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
a, 1+6+11+16+...+46+51
Số số hạng là : (51-1):5+1 = 11 ( số )
Tổng là : (51+1).11:2=286
b, Đặt A = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+\dfrac{5^2}{16.21}+\dfrac{5^2}{21.26}+\dfrac{5^2}{26.31 } \)
\(\dfrac{1}{5}A=\) \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\)
\(\dfrac{1}{5}A=\) \(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\)
\(\dfrac{1}{5}A=1-\dfrac{1}{31}\)
\(\dfrac{1}{5}A=\dfrac{30}{31}\)
\(A=\dfrac{30}{31}:\dfrac{1}{5}=\dfrac{150}{31}\)
Vậy..
\(B=\frac{5^2}{1.6}+\frac{5^2}{6.11}+.....+\frac{5^2}{26.31}\)
\(B=\frac{5.5}{1.6}+\frac{5.5}{6.11}+.....+\frac{5.5}{26.31}\)
\(B=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+.......+\frac{5}{26.31}\right)\)
\(B=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+......+\frac{1}{26}-\frac{1}{31}\right)\)
\(B=5.\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(B=\frac{5.30}{31}\)
\(B=\frac{150}{31}\)
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 \(A=\frac{150}{31}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)=\frac{5.30}{31}=\frac{150}{31}\)
#)Giải :
Ta có :
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)=5\left(1-\frac{1}{31}\right)=5\times\frac{30}{31}=\frac{150}{31}>1\)
\(\Rightarrow A>1\)