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ta co \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
=\(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
=\(5.\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}+\frac{26-21}{21.26}+\frac{31-26}{26.31}\right)\)
=\(5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
=\(5.\left(1-\frac{1}{31}\right)\)
=\(5.\frac{30}{31}\)
=\(\frac{150}{31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{6}{31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
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 ~
\(S=\frac{5^2}{1\cdot6}\cdot\frac{5^2}{6\cdot11}\cdot\frac{5^2}{11\cdot16}\cdot\frac{5^2}{16\cdot21}\cdot\frac{5^2}{21\cdot26}\)
\(=5\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{5}{16\cdot21}+\frac{5}{21\cdot26}\right)\)
\(=5\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)
\(=5\cdot\left(1-\frac{1}{26}\right)\)
\(=5\cdot\frac{25}{26}\)
\(=\frac{125}{26}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
Gọi biểu thức trên là A
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 \(\frac{3}{5}< \frac{3+m}{5+m}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(=5\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{150}{31}\)
\(1)\) \(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
\(2)\) Đặt \(A=4+2^2+2^4+...+2^{20}\)
\(4A=2^4+2^4+2^6+...+2^{22}\)
\(4A-A=\left(2^4+2^4+2^6+...+2^{22}\right)+\left(2^2+2^2+2^4+...+2^{20}\right)\)
\(3A=2^4+2^{22}-2^2-2^2\)
\(3A=2^{22}+2^4-2^3\)
\(A=\frac{2^{22}+2^4-2^3}{3}\)
\(3)\) \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\) ( bạn ghi đầy đủ ra nhé ở đây mk viết "..." cho nhanh )
\(=\)\(5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=\)\(5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\)\(5\left(1-\frac{1}{31}\right)\)
\(=\)\(5.\frac{30}{31}\)
\(=\)\(\frac{150}{31}\)
Chúc bạn học tốt ~
Ta có:
\(\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
B = \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+........+\frac{5^2}{26.31}\)
= \(5-\frac{5}{6}+\frac{5}{6}-\frac{5}{11}+\frac{5}{11}-\frac{5}{16}+......+\frac{5}{26}-\frac{5}{31}\)
= \(5-\frac{5}{31}=\frac{155-5}{31}=\frac{150}{31}\)
BẠn là ngừoi đẹp nhất thế giới nếu bạn tick cho mình^_- !!
#)Giải :
Ta có :
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)=5\left(1-\frac{1}{31}\right)=5\times\frac{30}{31}=\frac{150}{31}>1\)
\(\Rightarrow A>1\)
a) 2A= 1+1/2^2+1/2^3+...+1/2^2015+1/2^2016
2A-A=(1+1/2+1/2^2+...+1/2^2015+1/2^2016)-(1/2+1/2^2+...+1/2^2016+1/2^2017)
A= 1-1/2^2017
b) B=5.(5/1.6+5/6.11+...+5/26.31)
B=5.(1/5-1/6+1/6-1/11+1/11...-1/26+1/26-1/31)
B= 5.(1/5-1/31)
B=5.26/155
B=26/31
cần giải ko, đợi 1 chút 30p sau tôi giúp cho