Tính giá trị của biểu thức:
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
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`A = ( 5^2 )/( 1*6)+(5^2)/(6*11)+.....+(5^2)/(26*31)`
`= 5*( 5/( 1*6)+ 5/(6*11)+.....+5/(26*31))`
`= 5*( 1 - 1/6 + 1/6 - 1/11 +....+1/26 - 1/31 )`
`= 5*( 1 - 1/31 )`
`= 5 * 30/31 = 150/31`
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
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
\(B=\frac{5^2}{1.6}+\frac{5^2}{6.11}+.....+\frac{5^2}{26.31}\)
\(B=\frac{5.5}{1.6}+\frac{5.5}{6.11}+.....+\frac{5.5}{26.31}\)
\(B=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+.......+\frac{5}{26.31}\right)\)
\(B=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+......+\frac{1}{26}-\frac{1}{31}\right)\)
\(B=5.\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(B=\frac{5.30}{31}\)
\(B=\frac{150}{31}\)
\(C=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}+\frac{5}{31.35}\right).\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}+\frac{1}{31.35}\right)\)
\(=5\left(1-\frac{1}{31}+\frac{1}{31.35}\right)\)
\(=\frac{10501}{2107}\)
Ta có : \(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{31.36}\)
\(\Leftrightarrow\frac{C}{5}=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{31.36}\)
\(=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\)
\(=1-\frac{1}{36}=\frac{35}{36}\)
\(\Rightarrow C=\frac{35}{36}\cdot5=\frac{175}{36}\)
Vậy : \(C=\frac{175}{36}\)
P/s : Không chắc lắm :))
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 \(\frac{3}{5}< \frac{3+m}{5+m}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(=5\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{150}{31}\)
Vì: 52=5.5
= \(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
= \(5.\left(\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
= \(5.\left(1-\frac{1}{31}\right)\)
= \(5.\frac{30}{31}\)
= \(\frac{150}{31}\)
\(B=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
\(B=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(B=5\left(1-\frac{1}{31}\right)\)(TỐI GIẢN CÁC PHÂN SỐ GIỐNG NHAU)
\(B=5.\frac{30}{31}\)
\(B=\frac{150}{31}\)
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26+31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)
Vậy...........