3 – x trên 2009 – cho 2 – x trên 2010 – cho 1 – x trên 2011 bằng 1 27/10/2021 Bởi Reagan 3 – x trên 2009 – cho 2 – x trên 2010 – cho 1 – x trên 2011 bằng 1
Đáp án: Giải thích các bước giải: $\frac{3-x}{2009}$-$\frac{2-x}{2010}$ +$\frac{1-x}{2011}$=−1 →(1+$\frac{3-x}{2009}$ )-(1+$\frac{2-x}{2010}$ +(1+$\frac{1-x}{2011}$) =0 →$\frac{2009+3-x}{2009}$ -$\frac{2010+2-x}{2010}$ +$\frac{2011+1-x}{2011}$ =0 →$\frac{2012-x}{2009}$ -$\frac{2012-x}{2010}$ $\frac{2012-x}{2011}$ =0 →(2012-x)($\frac{1}{2009}$ -$\frac{1}{2010}$ +$\frac{1}{2011}$ )=0 →x=2012 Bình luận
$\dfrac{3-x}{2009}-\dfrac{2-x}{2010}-\dfrac{1-x}{2011}=1$ $\to \dfrac{3-x}{2009}+1-(\dfrac{2-x}{2010}+1)-(\dfrac{1-x}{2011}+1)=0$ $\to \dfrac{3-x+2019}{2009}-\dfrac{2-x+2010}{2010}-\dfrac{1-x+2011}{2011}=0$ $\to (2012-x)(\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011})=0$ $\text{Do:} \dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011} \ne 0$ $\to 2012-x=0$ $\to x=2012$ $\text{Vậy S={2012}}$ Bình luận
Đáp án:
Giải thích các bước giải:
$\frac{3-x}{2009}$-$\frac{2-x}{2010}$ +$\frac{1-x}{2011}$=−1
→(1+$\frac{3-x}{2009}$ )-(1+$\frac{2-x}{2010}$ +(1+$\frac{1-x}{2011}$) =0
→$\frac{2009+3-x}{2009}$ -$\frac{2010+2-x}{2010}$ +$\frac{2011+1-x}{2011}$ =0
→$\frac{2012-x}{2009}$ -$\frac{2012-x}{2010}$ $\frac{2012-x}{2011}$ =0
→(2012-x)($\frac{1}{2009}$ -$\frac{1}{2010}$ +$\frac{1}{2011}$ )=0
→x=2012
$\dfrac{3-x}{2009}-\dfrac{2-x}{2010}-\dfrac{1-x}{2011}=1$
$\to \dfrac{3-x}{2009}+1-(\dfrac{2-x}{2010}+1)-(\dfrac{1-x}{2011}+1)=0$
$\to \dfrac{3-x+2019}{2009}-\dfrac{2-x+2010}{2010}-\dfrac{1-x+2011}{2011}=0$
$\to (2012-x)(\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011})=0$
$\text{Do:} \dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011} \ne 0$
$\to 2012-x=0$
$\to x=2012$
$\text{Vậy S={2012}}$