Tính giá trị biểu thức:
C= \(b^3+c^3+ab^2+ac^2-abc\) biết a+b+c=0
H=\(\dfrac{1}{2003}.3\dfrac{1}{2005}.4\dfrac{2002}{2003}.\dfrac{4}{2005}-\dfrac{5}{2003.2004}+\dfrac{4}{401}\)
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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
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 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
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}-d...
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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
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
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b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
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Tính giá trị các biểu thức sau :
\(A=a.\dfrac{1}{2}+a.\dfrac{1}{3}-a.\dfrac{1}{4}\) với \(a=\dfrac{-4}{5}\)
\(B=\dfrac{3}{4}.b+\dfrac{4}{3}.b-\dfrac{1}{2}.b\) với \(b=\dfrac{6}{19}\)
\(C=c.\dfrac{3}{4}+c.\dfrac{5}{6}-c.\dfrac{19}{12}\) với \(c=\dfrac{2002}{2003}\)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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Lời giải
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm các thừa số chung a; b; c ra ngoài, sau đó tính phép tính trong ngoặc rồi thay giá trị a; b; c đã cho vào.
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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=?
5 tháng 3 2021 lúc 21:27
Giải phương trình :a,\(\dfrac{2x 1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)b,\(\dfrac{3\left(2x 1\right)}{4}-5-\dfrac{3x 2}{10}=\dfrac{2\left(3x-1\right)}{5}\)\(c,\dfrac{x 1}{2009} \dfrac{x 3}{2007}=\dfrac{x 5}{2005} \dfrac{x 7}{2003}\)\(d,\dfrac{392-x...
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a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
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Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
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a)
PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)
\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)
b)
PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)
\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)
Vậy pt vô nghiệm.
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Xem thêm câu trả lời
1, P 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{2}{2004}}
2, Q ( dfrac{1,5+1-0,75}{2,5+dfrac{5}{3}-1,25} + dfrac{0,375-0,3+dfrac{3}{11}+dfrac{3}{12}}{-0,625+0,5-dfrac{5}{11}-dfrac{5}{12}} ) : dfrac{1980}{3758} + 155
3, A 1.3 + 2.4 + 3.5 +....+ 97.99 + 98.100
4, B 1.2.3 + 2.3.4. +...+ 48.49.50
5, C dfrac{1}{1.2.3.4} + dfrac{1}{2.3.4.5} +...+ df...
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1, P = \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}\) - \(\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{2}{2004}}\)
2, Q = ( \(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\) + \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\) ) : \(\dfrac{1980}{3758}\) + 155
3, A = 1.3 + 2.4 + 3.5 +....+ 97.99 + 98.100
4, B = 1.2.3 + 2.3.4. +...+ 48.49.50
5, C = \(\dfrac{1}{1.2.3.4}\) + \(\dfrac{1}{2.3.4.5}\) +...+ \(\dfrac{1}{27.28.29.30}\)
6, D = 1 + \(2^2\) + \(2^4\) + \(2^6\) + .... +\(2^{200}\)
7, E = \(\dfrac{1}{3.5}\)+ \(\dfrac{5}{5.7}\) +...+ \(\dfrac{1}{97.99}\)
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
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Bài 1: Thực hiện phép tính:
a) ( 2- dfrac{3}{2}). ( 2- dfrac{4}{3}). (2- dfrac{5}{4}). ( 2- dfrac{6}{5})
b) dfrac{1}{2002}+dfrac{2003.2001}{2002}-2003
c)( dfrac{2003}{2004}+dfrac{2004}{2003}):dfrac{8028045}{8028024}
d) 4+ dfrac{1}{1+dfrac{1}{1+dfrac{2}{1+dfrac{3}{4}}}}
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Bài 1: Thực hiện phép tính:
a) ( 2- \(\dfrac{3}{2}\)). ( 2- \(\dfrac{4}{3}\)). (2- \(\dfrac{5}{4}\)). ( 2- \(\dfrac{6}{5}\))
b) \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)
c)( \(\dfrac{2003}{2004}+\dfrac{2004}{2003}\)):\(\dfrac{8028045}{8028024}\)
d) 4+ \(\dfrac{1}{1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{4}}}}\)