Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`x-(5/6 -x) =x-2/3`
`x-5/6 +x -x+2/3 =0`
`x = 5/6-2/3 = 5/6 -4/6 = 1/6`
`x+(x+1)+(x+2)+...+(x+30)=1240`
`=> (x + x + x + ... + x) + (1 + 2 + 3 +... + 30) = 1240`
`=> 31x + 465 = 1240`
`=> 31 x = 1240 - 465`
`⇒ 31x = 775`
`⇒ x = 775 : 31`
`⇒ x = 25`
\(\frac{1}{2013}.x+1+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{2012.2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+2-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x=\frac{1}{2013}\Rightarrow x=1\)
Vậy x=1
CHÚC CÁC EM HỌC TỐT
x+(x-1)+(x-2)+....+(x-50)=225
<=>x+x-1+x-2+...+x-50=225
<=>51x-1275=225
<=>51x=-1050
Lời giải:
Dãy $x,x+1, x+2,..., 2002$ có số số hạng là:
$\frac{2002-x}{1}+1=2003-x$
Tổng $x+(x+1)+....+2001+2002=\frac{(2002+x)(2003-x)}{2}$
Do đó:
$\frac{(2002+x)(2003-x)}{2}=2002$
$\Rightarrow (2002+x)(2003-x)=4004$
$2002.2003+x-x^2=4004$
$x^2-x-4006002=0$
$(x-2002)(x+2001)=0$
$\Rightarrow x=2002$ hoặc $x=-2001$
1/2013.x+1+1/2+1/6+1/12+...+1/2012.2013=2
1/2013.x+1+1/1.2+1/2.3+1/3.4+...+1/2012.2013=2
1/2013.x+1+1-1/2+1/2-1/3+1/3-1/4+...+1/2012-1/2013=2
1/2013.x+2-1/2013=2
1/2013.x =2-2+1/2013
1/2013.x =1/2013
=>2013.x=2013
=> x=1
\(\Rightarrow\frac{1}{2013.x}+1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2012}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013.x}+2-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013.x}=2-2+\frac{1}{2013}\)
\(\Rightarrow\frac{1}{2013.x}=\frac{1}{2013}\)
\(\Rightarrow2013.x=2013\)
\(\Rightarrow x=1\)
Ta có : x + (x + 1) + (x + 2) + .... + (x + 2012) = 2012.2013
<=> (x + x + x + ..... + x) + (1 + 2 + .... + 2012) = 2012.2013
<=> 2013x + \(\frac{2012.2013}{2}\) = 2012.2013
<=> 2013x = 2012.2013 - \(\frac{2012.2013}{2}\)
<=> 2013x = 2025078