Tính nhanh :
a) \(A=\dfrac{2004^3+1}{2004^2-2003}\)
b) \(B=\dfrac{2005^3-1}{2005^2+2006}\)
Những câu hỏi liên quan
Tính nhanh các kết quả sau:
\(A=\frac{2004^3+1}{2004^2-2003}\)
B=\(\frac{2005^3-1}{2005^2+2006}\)
\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{2004+1}{1-2003}\)\(=\frac{2005}{-2002}\)
\(B=\frac{2005^3-1}{2005^2+2006}\)\(=\frac{2005-1}{1+2006}=\frac{2004}{2007}\)
\(\Rightarrow A>B\)
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\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}\)
\(A=\frac{2005.\left(2004^2-2003\right)}{2004^2-2003}=2005\)
\(B=\frac{2005^3-1}{2005^2+2006}\)
\(B=\frac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=\frac{2004.\left(2005^2+2006\right)}{2005^2+2006}=2004\)
Tham khảo nhé~
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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 NHANH:
C= \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4} +....+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+....+\dfrac{1}{2006}}\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
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Tinh nhanh :
a) Tu so : 2005*2007-1
Mau so : 2004+2005*2006
b) Tu so : 2003*2004+2005*10+1994
Mau so: 2005*2004-2003*2004
a) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{\left(2014+1\right).2007-1}{2004+2005.2006}=\frac{2004+2005.2007-1}{2004+2005-2006}=\frac{2004+2005.2006}{2004+2005.2006}=1\)
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Cho : A = 1/1*2+1/3*4+1/5*6+.....+1/2003*2004+1/2005*2006
B=1/1004*2006+1/1005*2005+1/1006*2004+....+1/2006*1004
Tinh: A/B
Tính nhanh tổng đại số sau:
a) S=1-2-3+4+5-6-7+8+...+2001-2002-2003+2004
b) S=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006
a) Ta có: S = 1 - 2 - 3 + 4 + 5 - 6 - 7+ 8 + ... + 2001 - 2002 - 2003 + 2004
\(\Rightarrow\) S = (1 - 2 - 3 + 4) + (5 - 6 - 7+ 8) + ... + (2001 - 2002 - 2003 + 2004)
\(\Rightarrow\) S = (-4 + 4) + (-8 + 8) + ... + (-2004 + 2004)
\(\Rightarrow\) S = 0 + 0 + ... + 0
\(\Rightarrow\) S = 0
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a) 2005*2007-1 b)2003*2004+2005*10+1994
2004+2005*2006 2005*2004-2003*2004
c) ( 5+3/8+18+1/2-7-5/24 )
c) 5 + \(\frac{3}{8}\)+18 + \(\frac{1}{2}\) - 7 - \(\frac{5}{24}\)
=\(\frac{43}{8}\)+ \(\frac{35}{2}\) +\(\frac{163}{24}\)
=\(\frac{129}{24}\)+ \(\frac{420}{24}\)+\(\frac{163}{24}\)
= \(\frac{58}{51}\)
k nhé
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\(\frac{2005\times2007-1}{\left(2005+2005\times2006\right)-1}=\)\(\frac{2005\times2007-1}{2005\left(1+2006\right)-1}=\frac{2005\times2007-1}{2005\times2007-1}=1\)
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19 tháng 6 2019 lúc 8:05
Cho dfrac{a}{b} dfrac{c}{d} . Chứng minh :
a, dfrac{a^{2005}}{b^{2005}} dfrac{(a-c)^{2005}}{(b-d)^{2005}}
b, dfrac{(a^2+b^2)^3}{(c^2+d^2)^3} dfrac{a^3+b^3)^2}{(c^3+d^3)^2}
c, (dfrac{a-b}{c-d})^{2005} dfrac{2.a^{2005}-b^{2005}}{2.c^{2005}-d^{2005}}
d, dfrac{(a^2-b^2)^5}{(c^2-d^2)^5}
dfrac{a^{10}+b^{10}}{c^{10}+d^{10}}
e, dfrac{2.a^{2005}+5.b^{2005}}{2.c^{2005}+5.d^{2005}} dfrac{(a+b)^{2005}}{(c+d)^{2005}}
f, dfrac{(a^{2004}+b^{2004})^{2005}}{(c^{2004}+d^{2004})^{2005}} dfra...
Đọc tiếp
Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{a^{2005}}{b^{2005}} = \dfrac{(a-c)^{2005}}{(b-d)^{2005}}\)
b, \(\dfrac{(a^2+b^2)^3}{(c^2+d^2)^3}\) =\(\dfrac{a^3+b^3)^2}{(c^3+d^3)^2}\)
c, \((\dfrac{a-b}{c-d})^{2005}\) = \(\dfrac{2.a^{2005}-b^{2005}}{2.c^{2005}-d^{2005}}\)
d, \(\dfrac{(a^2-b^2)^5}{(c^2-d^2)^5} = \) \(\dfrac{a^{10}+b^{10}}{c^{10}+d^{10}}\)
e, \(\dfrac{2.a^{2005}+5.b^{2005}}{2.c^{2005}+5.d^{2005}}\) = \(\dfrac{(a+b)^{2005}}{(c+d)^{2005}}\)
f, \(\dfrac{(a^{2004}+b^{2004})^{2005}}{(c^{2004}+d^{2004})^{2005}}\) = \(\dfrac{(a^{2005} -b^{2005})^{2004}}{(c^{2005}-d^{2005})^{2004}}\)
cho hỏi chút
\(\frac{a}{b}=\frac{c}{d}\)
trong đó
\(a=c\) hay \(a\ne c\)
\(b=d\) hay \(b\ne d\)
( bài có thiếu điều kiện ko vậy )
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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}...
Đọc tiếp
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
