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a. 2006/2005 x 2007/2006 x 2008/2007 x 2009/2008 x 2010/2009'
= 2006 x 2007 x 2008 x 2009 x 2010 / 2005 x 2006 x 2007 x 2008 x 2009
= 2010/2005
= 402/401
\(\left(1+\frac{1}{2005}\right)x\left(1+\frac{1}{2006}\right)x\left(1+\frac{1}{2007}\right)x\left(1+\frac{1}{2008}\right)x\left(1+\frac{1}{2009}\right)\)
\(=\frac{2006}{2005}x\frac{2007}{2006}x\frac{2008}{2007}x\frac{2009}{2008}x\frac{2010}{2009}\)
\(=\frac{2010}{2005}\)
\(=\frac{402}{401}\)
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{8}\right):15\frac{3}{2}=2008\)
\(\Leftrightarrow2009-\left(\frac{41}{9}+x-\frac{63}{8}\right):\frac{33}{2}=2008\)
\(\Leftrightarrow2009-\left(x-\frac{239}{72}\right):\frac{33}{2}=2008\)
\(\Leftrightarrow2009-\frac{2x}{33}+\frac{239}{1188}=2008\)
\(\Leftrightarrow\frac{-2x}{33}=\frac{-1427}{1188}\)
\(\Leftrightarrow-2376x=-47091\)
\(\Leftrightarrow x=\frac{1427}{72}\)
\(A=\frac{2007\cdot\left(2008-1008\right)}{\left(2007-1007\right)+\left(2008-1008\right)}=\frac{2007\cdot1000}{1000+1000}=\frac{2007}{2}\)
\(B=\frac{1978\cdot1979+\left(1979+1\right)\cdot21+\left(1979-21\right)}{1979\cdot\left(1980-1978\right)}=\frac{1979\cdot\left(1978+21\right)}{1979\cdot2}=\frac{1999}{2}\)
Mình không tin đây là toán lớp 5 đâu ! ( TẠI VÌ MÌNH CHƯA HỌC ) ^_^
c) (x+1) + (x+2) + ... + (x+5) = 90
=> 5x + ( 1 + 2 + ... + 5 ) = 90
5x + 15 = 90
5x = 90 - 15
5x = 75
x = 75 : 5
x = 15
d) (x+1) + (x+2) + .... + (x+100) = 20150
=> 100x + ( 1+2+...+100 ) = 20150
100x + 5050 = 20150
100x = 20150 - 5050
100x = 15100
x = 15100 : 100
x = 151
Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90
<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90
<=> 5x + 15 = 90
=> 5x = 75
=> x = 15
( x . 25 + 2008 ) . 2009 = ( 575 + 2008 ) . 2009
=> x . 25 + 2008 = 575 + 2008
=> x . 25 = 575
x = 575 : 25
x = 23
Dấu " . " là dấu nhân nha
=>x . 25 +2008 =575+2008
=>x.25=575
x=575:5
x=115