2009 - ( \(\frac{41}{9}\)+ x - \(\frac{133}{18}\)) : \(\frac{47}{3}\)= 2008
2009 - ( 41/9+ x - 133/18) : 47/3= 2008
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 = \(\frac{37}{2}\)
Tìm x ,biết: 2009 - ( 41/9 + x - 133/18 ) : 47/3 = 2008
Các bạn (anh,chị) giải dùm mình (em) với.
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 - (41/9 + x - 133/18) : 47/3 = 2008
Mình làm rồi 111/6 nhưng mình đem cho cô giáo thì cô lại bảo sai
Giúp mình nhé nhanh,đúng và đầy đủ mình tick
tìm x:
2009 - ( 4\(\frac{5}{9}\)+ x - 7\(\frac{7}{18}\)) : 15\(\frac{2}{3}\)= 2008
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right):15\frac{2}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2008\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2009-2008\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=1\)
\(\frac{41}{9}+x-\frac{133}{18}=1\cdot\frac{47}{3}\)
\(\frac{41}{9}+x-\frac{133}{18}=\frac{47}{3}\)
\(\frac{41}{9}+x=\frac{133}{18}+\frac{47}{3}\)
\(\frac{41}{9}+x=\frac{415}{18}\)
\(x=\frac{415}{18}-\frac{41}{9}\)
\(x=\frac{37}{2}\)
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right)\div15\frac{2}{3}=2008\)
\(\left(4\frac{5}{9}+x-7\frac{7}{18}\right)\div15\frac{2}{3}=2009-2008\)
\(\left(-3\frac{3}{18}+x\right)\div15\frac{2}{3}=1\)
\(-3\frac{3}{18}+x=15\frac{2}{3}\)
\(x=15\frac{2}{3}+3\frac{3}{18}\)
\(x=15\frac{12}{18}+3\frac{3}{18}\)
\(x=18\frac{15}{18}\)
exactly 100%
- Giải phương trình: \(\frac{x-2009-2010}{2008}+\frac{x-2008-2010}{2009}+\frac{x-2008-2009}{2010}=3\)
\(\frac{x-2009-2010}{2008}+\frac{x-2008-2010}{2009}+\frac{x-2008-2009}{2010}=3\)
\(\Leftrightarrow\frac{x-2008-2009-2010}{2008}+\frac{x-2008-2009-2010}{2009}+\frac{x-2008-2009-2010}{2010}=0\)
\(\Leftrightarrow\left(x-2008-2009-2010\right)\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
\(\Leftrightarrow x-6027=0\Leftrightarrow x=6027\)
\((x-1)/2013+(x-2)/2012+(x-3)/2011+(x-4)/2010+(x-5)/2009+(x-6)/2008\)
\(\frac{1}{x^2+9\cdot x+20}+\frac{1}{x^2+11\cdot x+30}+\frac{1}{x^2+13\cdot x+42}=\frac{1}{18}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\) ( có lẽ đề như này )
\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)
\(\Leftrightarrow x=2014\)
...
Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)
\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow3.18=x^2+4x+7x+28\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
tính tổng sau :\(c=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\)\(\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
Tính nhanh:
54 x 113 + 45 x 113 + 113
54 x 47 - 47 x 53 - 20 - 27
10000 - 47 x 72 - 47 x 28
(145 x 99 + 145) - (143 x 101 - 143)
1002 x 9 - 18
8 x 427 x 3 + 6 x 573 x 4
2008 x 867 + 2009 x 133
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
a, 10000 - 47 x 72 x 28 =
b, ( 145 x 99 + 145 ) - ( 143 x 101 - 143 ) =
c, 1002 x 9 - 18 =
d, 8 x 427 x 3 + 6 x 573 x 4 =
e, 2008 x 867 x 2009 x 133 =