Question

Tìm x:

x+1/10 + x+1/11 + x+1/12 = x+1/13 + x+1/14

x+4/2000 + x+3/2001 = x+2/2002 + x+1/2003

Answer 1

$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

= $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}$

$\Rightarrow \left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)$

Vì 10<11<12<13<14 $\Rightarrow \frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}$

$\Rightarrow \frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$

$\Rightarrow x+1=0$

$\Rightarrow x=-1$

Answer 2

 Câu 1:x+1/10 + x+1/11 = x+1/12 + x+1/13 + x+1/14.

<-> (x+1)(1/10+1/11-1/12-1/13-1/14)=0

<-> x+1=0

<-> x=-1

Câu 2:

x+4/2000+x+3/2001=x+2/2002+x

⇔x+4/2000+1+x+3/2001=x+2/2002+1+x+1/2003

⇔x+2004/2000+x+2004/2001=x+2004/2002+x+2004/2003

⇔(x+2004)/(1/2000+1/2001−1/2002−1/2003)=0

⇔x+2004=0

⇔x=-2004