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Đặt \(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(\Rightarrow A=\frac{5^2}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
\(\Rightarrow A=5.\left(1-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)
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 \(A=\frac{150}{31}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)=\frac{5.30}{31}=\frac{150}{31}\)
A=\(\frac{5^2}{1.6}\)+\(\frac{5^2}{6.11}\)+....+\(\frac{5^2}{26.31}\)=\(\frac{25}{1.6}\)+\(\frac{25}{6.11}\)+.....+\(\frac{25}{26.31}\)
\(\frac{1}{5}\)A=\(\frac{5}{1.6}\)+\(\frac{5}{6.11}\)+....+\(\frac{5}{26.31}\)=1-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{11}\)+....+\(\frac{1}{26}\)-\(\frac{1}{31}\)=1-\(\frac{1}{31}\)=\(\frac{30}{31}\)
A=\(\frac{30}{31}\):\(\frac{1}{5}\)
A=\(\frac{150}{31}\)
Câu 1:
Giả sử \(\frac{3}{5}< \frac{3+m}{5+m}\)
=) \(3.\left(5+m\right)< 5.\left(3+m\right)\)
=) \(15+3m< 15+5m\) ( Đúng vì \(15=15\)và \(3m< 5m\)) =) Điều giả sử đúng
=) \(\frac{3}{5}< \frac{3+m}{5+m}\)
* Từ điều trên ta suy ra : Nếu \(\frac{a}{b}< 1\)=) \(\frac{a}{b}< \frac{a+m}{b+m}\)
Và nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Câu 2 :
= \(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{31}\right)\)= \(5.\frac{30}{31}=\frac{150}{31}\)
=> Với mọi số tự nhiên m ( như m\(\ne\)0 ) thì \(\frac{3}{5}< \frac{3+m}{5+m}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(=5\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{150}{31}\)
ta co \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
=\(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
=\(5.\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}+\frac{26-21}{21.26}+\frac{31-26}{26.31}\right)\)
=\(5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
=\(5.\left(1-\frac{1}{31}\right)\)
=\(5.\frac{30}{31}\)
=\(\frac{150}{31}\)
Ta có :
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)
\(S=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)
\(S=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)
\(S=5\left(1-\frac{1}{26}\right)\)
\(S=5.\frac{25}{26}\)
\(S=\frac{125}{26}\)
Vậy \(S=\frac{125}{26}\)
Chúc bạn học tốt ~
a) áp dụng dãy số cách đều đi
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..