2009 - ( 41/9+ x - 133/18) : 47/3= 2008
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đề....
(41/9+x-133/18):47/3=2009-2008
(41/9+x-133/18):47/3=1
(41/9+x-133/18)=1*47/3=47/3
41/9+x=47/3+133/18
41/9+x=415/18
x=415/18-41/9
x=37/2
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=2008\cdot\frac{47}{3}\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=\frac{94376}{3}\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=2009-\frac{94376}{3}\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=-\frac{88349}{3}\)
\(\frac{41}{9}+x\) \(=-\frac{88349}{3}+\frac{133}{18}\)
\(\frac{41}{9}+x\) \(=-\frac{529961}{18}\)
\(x=-\frac{529961}{18}-\frac{41}{9}\)
\(x=-\frac{176681}{6}\)
a/ = 113 x ( 54 + 45 + 1 ) = 113 x 100 = 11300
b/ = 47 x ( 54 - 53 ) - 20 - 27 = 47 - 20 - 27 = 0
c/ = 145 x ( 99 + 1 ) - 143 x ( 101 - 1 ) = 145 x 100 - 143 x 100 = 100 x ( 145 - 143 ) = 100 x 2 = 200
d/ = 1002 x 9 - 2 x 9 = 9 x ( 1002 - 2 ) = 9 x 1000 = 9000
e/ = 24 x 427 + 24 x 573 = 24 x ( 427 + 573 ) = 24 x 1000 = 24000
f/ Sửa đề bài là :
2008 x 867 + 2008 x 133 = 2008 x ( 867 + 133 ) = 2008 x 1000 = 2008000
3457 - 27 x 48 - 48 x 73 + 6543
=3457 - 48 x ( 27 + 73) + 6543
=3457 - 48 x 100 + 6543
=(3457+ 6543) - 4800
=10000- 4800
=5200
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
8 x 427 x 3 + 6 x 573 x 4
= 8 x 3 x 427 + 6 x 4 x 573
= 24 x 427 + 24 x 573
= 24 x (427 + 573)
= 24 x 1000
= 24000
2008 x 867 + 2009 x 133
= 2008 x 867 + 2008 x 133 + 133
= 2008 x (867 + 133) + 133
= 2008 x 1000 + 133
= 2008133
2009 - ( 41/9+ x - 133/18) : 47/3= 2008
( 41/9 + x - 133/18 ) : 47/3 = 2009 - 2008
( 41/9 + x - 133/18 ) : 47/3 = 1
41/9 + x - 133/18 = 1 x 47/3
41/9 + x - 133/18 = 47/3
41/9 + x = 47/3 + 133/18
41/9 + x = 415/18
x = 415/18 - 41/9
x = \(\frac{37}{2}\)