Tìm x: |x+2015| + |x+2016| + |x+2017| = 6x
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Điều kiện x \(\ge0\)từ đó ta có
x + 2015 + x + 2016 + x + 2017 = 6x
<=> x = 2016
2016 x 2016 - 2015 x 2017 + x = 2016
(2016 x 2016) - (2015 x 2017) + x = 2016
4064256 - 4064255 + x = 2016
1 + x = 2016
x = 2016 - 1
x = 1
Ta có:
x biết: 2016 x 2016 - 2015 x 2017 + x = 2016
x = 2015 x 2017 + 2016 - 2016 x 2016
x = 2015 x 2017 + 2016 x (1 - 2016)
x = 2015 x 2017 - 2015 x 2016
x = 2015 x (2017 - 2016)
x = 2015 x 1
x = 2015
\(\frac{x+2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)
\(\Leftrightarrow\frac{x+2015}{2016}+1+\frac{x+2016}{2015}+1+\frac{x+2017}{2014}+1=0\)
\(\Leftrightarrow\frac{x+4031}{2016}+\frac{x+4031}{2015}+\frac{x+4031}{2014}=0\)
\(\Leftrightarrow\left(x+4031\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
\(\Rightarrow x+4031=0\)
\(\Rightarrow x=-4031\)
Ta có \(\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|\ge0\Rightarrow6x\ge0\Rightarrow x\ge0\)
=> \(\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|=3x+6048=6x\Rightarrow3x=6048\Rightarrow x=2016\)
Vậy x=2016
Ta có:
\(\left|x+2015\right|\ge0\)
\(\left|x+2016\right|\ge0\)
\(\left|x+2017\right|\ge0\)
\(\Rightarrow\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x\right|=x\)
\(\Rightarrow\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|=\left(x+2015\right)+\left(x+2016\right)+\left(x+2017\right)=6x\)
\(\Rightarrow3x+6048=6x\)
\(\Rightarrow3x=6048\)
\(\Rightarrow x=2016\)
Vậy \(x=2016\)