Tìm x:
a) x + 3.x + 5.x + .... + 2009.x = 2010. 1005
b) x + (x+1) + (x+2) + .... + (x+30) = 620
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27 tháng 1 2023 lúc 0:59
Tìm x:
a, (x - 1/2) : 1/3 + 5/7 = 9 5/7
b, x + 30% x = - 1,31
c, -2/3.x + 1/5 = 1/10
\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)
\(\Leftrightarrow x-\dfrac{1}{3}=3\)
\(\Leftrightarrow x=3+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
\(b,x+30\%x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)
\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)
\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)
\(\Leftrightarrow x=\dfrac{3}{20}\)
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tìm x
x + ( x + 1 ) + ( x + 2 ) + ( x +3 ) + .......... + ( x + 2009 ) = 2009 . 2010
x+(x+1)+...+(x+2009) dãy số trên ta có : x+2009-x+1=2010
=> x+(x+1)+....+(x+2009)=(x+x+2009).2010:2=(2x+2009).1005
=>(2x+2009).1005= 2009.2010
2x+2009 = 2009.2
2x=2009.2-2009
2x=2009
x=\(\frac{2009}{2}\)
Tìm x biết: (x+1/2013) + (x+2/2012) + (x+3/2011) = (x+4/2010) + (x+5/2009) + (x+6/2008)
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
tìm x biết :
(2009 - x^2) + ( 2009 - x^2)( x - 2010) + ( x - 2010) / (2009 - x^2) - ( 2009 - x^2)( x - 2010) + ( x - 2010 ) = 19/49
Tìm x : (2009-x)2+(2009-x)(x-2010)+(x-2010)2/(2009-x)2-(2009-x)(x-2010)+(x-2010)2=19/49
TÌM X
{[2009-x]^2+[2009-x][x-2010]+[x-2010]^2}/{[2009-x]^2-[2009-x][x-2010]+[x-2010]^2}=19/49
Tìm x biết: (2009-x)2 + (2009-x)(x-2010) + (x-2010)2/(2009-x)2 - (2009-x)(x-2010) + (x-2010)2 = 19/49
dat a =2009-x
b=x-2010
ta co : a^2+ab+b^2/a^2-ab+b^2 =19/49
<=>49a^2+49ab+49b^2=19a^2-19a+19b^2
<=>30a^2+68a+30b^2=0
<=>15a^2+34ab+15b^2=0
<=>15a^2+9ab+25ab+15b^2=0
<=>3a(5a+3b)+5b(5a+3b)=0
<=>(5a+3b)(3a+5b)=0
<=>5a+3b=0 hoac 3a+5b=0
vs 5a +3b=0 <=>5(2009-x)+3(x-2010)=0=>x=......
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Tìm x biết: (2009-x)2 + (2009-x)(x-2010) + (x-2010)2/(2009-x)2 - (2009-x)(x-2010) + (x-2010)2 = 19/49
Tìm x thỏa mãn: x + (x + 1) + (x + 2) + … + 2009 + 2010 = 2010 A.-2010 B.-2008 C.0 D.-2009
