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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I
1
Y
27 tháng 1 2023
\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)
\(\Leftrightarrow x-\dfrac{1}{3}=3\)
\(\Leftrightarrow x=3+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
\(b,x+30\%x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)
\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)
\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)
\(\Leftrightarrow x=\dfrac{3}{20}\)
PL
1
SS
30 tháng 8 2021
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=\(\frac{2009}{2}\)
LT
1
YN
2 tháng 3 2022
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
HL
0
5 tháng 12 2018
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=......
a) x + 3.x + 5.x + ... + 2009. x = 2010.1005
\(x\times\left(1+3+5+...+2009\right)=2010\times1005\)
\(x\times\left[\left(1+2009\right)\times1005:2\right]=2010\times1005\)
\(x\times2010\times1005\times\frac{1}{2}=2010\times1005\)
\(\Rightarrow x\times\frac{1}{2}=2010\times1005:\left(2010\times1005\right)\)
\(x\times\frac{1}{2}=1\)
x = 2
b) x + (x+1) + (x+2) +...+ (x+30) = 620
x. 31 + ( 1+2+...+30) = 620
x.31 + [ ( 30+1).30:2) = 620
x.31 + 465 = 620
x.31 = 620 - 465
x.31 = 155
x = 155 : 31
x = 5
a) x+3.x+5.x+.....+2009.x = 2010.1005
=> x.(1+3+5+....+2009) = 2010.1005
=> x.1010025 = 2020050
=> x = 2
Vậy x = 2
b) x+(x+1)+(x+2)+....+(x+30) = 620
=> (x+x+x+...+x)+(1+2+3+...+30) = 620
=> 31x + 465 = 620
=> 31x = 155
=> x = 5