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\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+..+\dfrac{5}{26.31}\right)\)
\(A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(A=5\left(1-\dfrac{1}{31}\right)\)
\(A=5-\dfrac{1}{155}\)
\(A< 5\rightarrowđpcm\)
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+.......\dfrac{5^2}{26.31}\)
\(\Leftrightarrow A=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+.........+\dfrac{5}{26.31}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+..........+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{31}\right)\)
\(\Leftrightarrow A=5.\dfrac{30}{31}=\dfrac{150}{31}\)
a)
=\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5\left(7^3-7^4\right)}{7^3.3^2}\)
=\(\frac{3^4\left(3-1\right)}{^{ }3^4\left(9+3\right)}-\frac{5.7^3-5.7^4}{7^3.3^2}\)
=\(\frac{1}{6}-\frac{7^3.5\left(1-7\right)}{7^3.3^2}=\frac{1}{6}-\frac{30}{9}=-\frac{19}{6}\)
Vậy A=\(-\frac{19}{6}\)
câu b lúc nã mk làm sai rui
dây mới đúng
=\(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
=\(\frac{1}{5}\left(1-\frac{1}{101}\right)=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
a)
\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)
\(\Rightarrow A=\frac{17}{35}\)
b)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)
c)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)
a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(=\frac{1}{5}+\frac{2}{7}\)
\(=\frac{7}{35}+\frac{10}{35}\)
\(=\frac{17}{35}\)
Vậy \(A=\frac{17}{35}\)
b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)
\(=5.\frac{50}{671}\)
\(=\frac{250}{671}\)
Vậy \(B=\frac{250}{671}\)
\(A=\frac{10^2}{1\cdot6}+\frac{10^2}{6\cdot11}+...+\frac{10^2}{61\cdot66}=\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{61\cdot66}\right)\cdot20\)
\(=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{61}-\frac{1}{66}\right)\cdot20\)
\(=\left[\left(1-\frac{1}{66}\right)+\left(\frac{1}{6}-\frac{1}{6}\right)+...+\left(\frac{1}{61}-\frac{1}{61}\right)\right]\cdot20\)
\(=\left[\left(\frac{66}{66}-\frac{1}{66}\right)+0+...+0\right]\cdot20=\frac{65}{66}\cdot20=\frac{65\cdot20}{66}=\frac{65\cdot10}{33}=\frac{650}{33}\)
\(A=\frac{10^2}{1.6}+\frac{10^2}{6.11}+...+\frac{10^2}{61.66}\)
\(=10^2.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{61.66}\right)\)
\(=10^2.5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{61}-\frac{1}{66}\right)\)
\(=500.\left(1-\frac{1}{66}\right)\)
\(=500.\frac{65}{66}\)
\(=\frac{16250}{33}\)
A=3 /1^2.2^2 +5 / 2^2.3^2 +7/3^2.4^2 +...+ 19 /9^2.10^2
=1/1^2-1/2^2+1/2^2-1/3^2+1/3^2-1/4^2+....+1/9^2-1/10^2
=1/1^2-1/10^2
=99/100
=0,99
vậy A< 1
ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101
= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101
=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)
=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)
=1/5 . (1/1-1/101)
=1/5 . 100/101
= 20/101
5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)
=\(1- {1 \over 101}={100 \over 101}\)
suy ra A =\({20 \over 101}\)
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
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)