Thực hiện phép tính:
\((\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}).(\frac{1}{3}-\frac{1}{4}-\frac{1}{12})\)
Những câu hỏi liên quan
Thực hiện phép tính:
\((\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968})\times(\frac{1}{3}-\frac{1}{4}-\frac{1}{12})\)
GIÚP MÌNH VỚI Ạ. MÌNH ĐANG CẦN GẤP!
=(1975/1976+2010/2011+1963/1968)x(4/12-3/12-1/12)
=(1975/1976+2010/2011+1963/1968)x0
=0
bài 2
a) ( 1975/1976 + 2010/2011 + 1963/1968 ) : ( 1/3 - 1/4 - 1/12)
Sửa đề:
\(\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\dfrac{4-3-1}{12}\)
\(=\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\dfrac{0}{12}\)
\(=0\)
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1)thực hiện phép tính hợp lí nhất có thể:
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\right)\)
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(\Rightarrow D=\frac{1}{2012}\)
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Thực hiện phép tính bằng cách hợp lý:
A= \(\frac{2011.2010-1}{2009.2011+2010}\)
B= (\(\frac{1}{2009}+\frac{13}{2010}+\frac{135}{2011}\)).(\(\frac{1}{3}-\frac{1}{5}-\frac{2}{15}\))
tính nhanh
\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{7}}{\frac{5}{12}+1-\frac{7}{11}}\)
thực hiện phép tính :
1-2+3-4+5-6+...+2011-2012
1-2+3-4+5-6+...+2011-2012
=2012-2011+...+6-5+4-3+2-1
=(2012-2001)+...+(6-5)+(4-3)+(2-1)
=1+1+1+...+1+1(có 1006 số 1)
=1x60
=60
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Tính
a)\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....\frac{1}{2011}}\)
b)\(\frac{0.375-0.3+\frac{3}{11}+\frac{3}{12}}{-0.265+0.5-\frac{5}{11}-\frac{5}{12}}+\frac{1.5+1-0.75}{2.5+\frac{5}{3}-1.25}\)
Bài 3 : a) Tính
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) Tính :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\frac{1}{2011}}\)
Thực hiện phép tính :
a ) A =\(\frac{2015+2013+2011+2009+...+7+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
b) B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1
= [(2015 - 1) : 2 + 1].(2015 + 1) : 2
= 1008.2016 : 2 = 1016064
Lại có : 2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng
= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)
= 2 + 2 + ... + 2 + 2 (504 số hạng 2)
= 2 x 504 = 1008
Khi đó A = \(\frac{1016064}{1008}=1008\)
b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)
Khi đó : B/100 = 1/2
=> B = 50
Vậy B = 50
TÍNH : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}}\)
nâng cao phát triển toán 7 đấy
mấy bài đấu thì phải
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Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)
\(L=1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(L=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}\)
\(L=2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2012}\right)\)
Hay: \(P=\frac{1}{2012}\)
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