So sánh các số hữu tỉ sau bằng cách nhanh nhất
\(\dfrac{-1}{540}\) và \(\dfrac{1}{3780}\) \(\dfrac{2003}{-2004}\) và \(\dfrac{-2005}{2003}\)
\(\dfrac{-29}{202}\) và \(\dfrac{13}{-92}\)
Mk đang cần gấp
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so sánh M,N
M=\(\dfrac{2003}{2004}+\dfrac{2004}{2005}\)
N=\(\dfrac{2003+2004}{2004+2005}\)
Ta có:
N=\(\dfrac{2003+2004}{2004+2005}\)=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Ta thấy:
\(\dfrac{2003}{2004+2005}\)<\(\dfrac{2003}{2004}\)(1)
\(\dfrac{2004}{2004+2005}\)<\(\dfrac{2004}{2005}\)(2)
Từ (1) và (2) --> M=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\)>\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)=N
Vậy M>N
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so sánh A= \(\dfrac{2003^{2003}+1}{2003^{2004}+1}\)
B=
\(\dfrac{2003^{2002}+1}{2003^{2003}+1}\)
Ta có: \(2003^{2003}+1=2003^{2002+1}+1và2003^{2004}+1=2003^{2003+1}+1\)
\(\Rightarrow A>B\)
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so sánh
\(\dfrac{2003}{2004}+\dfrac{2004}{2005}\) và \(\dfrac{2003+2004}{2004+2005}\)
Cho A=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\); B=\(\dfrac{2003+2004}{2004+2005}\)
Ta có: B=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Vì: \(\dfrac{2003}{2004+2005}< \dfrac{2003}{2004}\)
\(\dfrac{2004}{2004+2005}< \dfrac{2004}{2005}\)
=>\(\dfrac{2003}{2004+2005}+\dfrac{2004}{2004+2004}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>\(\dfrac{2003+2004}{2004+2005}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>B<A
Vậy B<A
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I, Tìm x: a, \(\dfrac{x-2004}{2003}+\dfrac{x-2003}{2005}+\dfrac{x-2005}{2004}=3+\dfrac{2005}{2004}+\dfrac{2004}{2005}\)
Tính giá trị biểu thức:
Ddfrac{dfrac{1}{2003}+dfrac{1}{2004}+dfrac{1}{2005}}{dfrac{5}{2003}+dfrac{5}{2004}+dfrac{5}{2005}}-dfrac{dfrac{2}{2002}+dfrac{2}{2003}+dfrac{2}{2004}}{dfrac{2}{2002}+dfrac{3}{2003}+dfrac{3}{2004}}
Hdfrac{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2012}}{dfrac{2011}{1}+dfrac{2010}{2}+...+dfrac{1}{2011}}
Idfrac{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2012}}{dfrac{2012}{2}+dfrac{2012}{3}+...+dfrac{2012}{2011}}
Help me!
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Tính giá trị biểu thức:
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{2}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(H=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}}\)
\(I=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2012}{2}+\dfrac{2012}{3}+...+\dfrac{2012}{2011}}\)
Help me!
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\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)
\(D=\dfrac{1}{5}-\dfrac{2}{3}\)
\(D=-\dfrac{7}{15}\)
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Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
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Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tìm số hữu tỉ , biết rằng: \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+1-\dfrac{x}{2004}\)
\(\Leftrightarrow\left(\dfrac{2-x}{2002}+1\right)=\left(\dfrac{1-x}{2003}+1\right)+\left(1-\dfrac{x}{2004}\right)\)
=>2004-x=0
=>x=2004
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Adfrac{1}{4.9}+dfrac{1}{9.14}+dfrac{1}{14.19}+...+dfrac{1}{44.49}+left(dfrac{1-3-5-7-...-49}{89}right)
Bdfrac{212.3^5.4^6.9^2}{left(2^2.3right)^6+8^4.3^5}-dfrac{5^{10}.7^3-25^4.49^2}{left(125.71^3+59.14^3right)}
Cdfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}{2005}}{dfrac{5}{2003}+dfrac{5}{2004}-dfrac{5}{2005}}-dfrac{dfrac{2}{2002}+dfrac{2}{2003}-dfrac{2}{2004}}{dfrac{3}{2002}+dfrac{3}{2003}-dfrac{3}{2004}}
Dleft(dfrac{1,5+1-0,75}{2,5+dfrac{5}{3}-1,25}right)+left(dfrac{0,375-0,3+dfrac{3}{11}+df...
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A=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}+\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^2}{\left(125.71^3+59.14^3\right)}\)
C=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
D=\(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\right)+\left(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)
E=13+23+...+103=3025
Tính F=23+42+63+...+203=?
Tính giá trị của các biểu thức sau1) A1+2+2^2+...+2^{2015}2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right)3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89}4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3}5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}{2005...
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Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203