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\(\frac{x+5}{2017}+\frac{x+6}{2016}+\frac{x+7}{2015}=-3\)
\(\left(\frac{x+5}{2017}+1\right)+\left(\frac{x+6}{2016}+1\right)+\left(\frac{x+7}{2015}+1\right)=0\)
\(\frac{x+2022}{2017}+\frac{x+2022}{2016}+\frac{x+2022}{2015}=0\)
\(\left(x+2022\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)
\(x+2022=0\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\ne0\right)\)
x=-2022
Ta có:
|x−2015|+|x−2016|+|x−2017||x−2015|+|x−2016|+|x−2017|
=|x−2016|+|x−2015|+|x−2017|=|x−2016|+|x−2015|+|x−2017|
=|x−2016|+(|x−2015|+|x−2017|)=|x−2016|+(|x−2015|+|x−2017|)
∗)∗) Áp dụng BĐT |a|+|b|≥|a+b||a|+|b|≥|a+b| ta có:
|x−2015|+|x−2017|=|x−2015|+|x−2017|= |x−2015|+|2017−x||x−2015|+|2017−x|
≥|x−2015+2017−x|=|2|=2≥|x−2015+2017−x|=|2|=2
∗)∗) Dễ thấy: |x−2016|≥0∀x|x−2016|≥0∀x
⇔|x−2015|+|x−2016|+|x−2017|⇔|x−2015|+|x−2016|+|x−2017| ≥2≥2
Đẳng thức xảy ra ⇔⎧⎩⎨⎪⎪x−2015≥0x−2016=0x−2017≤0⇔⎧⎩⎨⎪⎪x≥2015x=2016x≤2017⇔{x−2015≥0x−2016=0x−2017≤0⇔{x≥2015x=2016x≤2017 ⇔x=2016⇔x=2016
Vậy GTNNGTNN của biểu thức là 2⇔x=2016
P= |x-2015|+|2016-x| +|x-2017|
=> P = |x-2015|+|x-2016| +|2017-x|
Ta có\(\left|x-2015\right|\ge x-2015\)(với mọi x)
\(\left|x-2016\right|\ge x-2016\)(với mọi x)
\(\left|x-2017\right|\ge x-2017\)(với mọi x)
\(\Rightarrow\left|x-2015\right|+\left|x-2016\right|+\left|x-2017\right|\ge x-2015+0+x-2017\)(với mọi x)
\(\Rightarrow P\ge2\)(với mọi x)
=> P đạt GTNN là 2 khi
\(\hept{\begin{cases}\left|x-2015\right|=0\\\left|x-2016\right|=0\\\left|x-2017\right|=0\end{cases}\hept{\begin{cases}x-2015\ge0\\x-2016=0\\x-2017\ge0\end{cases}\hept{\begin{cases}x\ge2015\\x=2016\\x\ge2017\end{cases}\Rightarrow}}x=2016}\)
Vậy GTNN của P là 2 tại x = 2016
Lời giải:
Ta có:
\(B=|x-2016|+|x-1|+2=|x-2016|+|1-x|+2\)
\(\geq |x-2016+1-x|+2=|-2015|+2=2017\)
Vậy \(B_{\min}=2017\)
Dấu "=" xảy ra khi \((x-2016)(1-x)\geq 0\Leftrightarrow 1\leq x\leq 2016\)
\(-\left(x+\dfrac{1}{8}\right)^{2016}-\left|y+5\right|-\left(x+z\right)^{2018}\)
Với mọi \(x;y;z\in R\) ta có:
\(\left\{{}\begin{matrix}-\left(x+\dfrac{1}{8}\right)^{2016}\le0\\-\left|y+5\right|\le0\\-\left(x+z\right)^{2018}\le0\end{matrix}\right.\)
\(\Rightarrow-\left(x+\dfrac{1}{8}\right)^{2016}-\left|y+5\right|-\left(x+z\right)^{2018}\le0\)
Ta có pt:
\(\left\{{}\begin{matrix}-\left(x+\dfrac{1}{8}\right)^{2016}-\left|y+5\right|-\left(x+z\right)^{2018}\ge0\\-\left(x+\dfrac{1}{8}\right)^{2016}-\left|y+5\right|-\left(x+z\right)^{2018}\le0\end{matrix}\right.\)
Nên \(-\left(x+\dfrac{1}{8}\right)^{2016}-\left|y+5\right|-\left(x+z\right)^{2018}=0\)
Nên cặp số \(x;y;z\) thỏa mãn là :\(\left\{{}\begin{matrix}x=-\dfrac{1}{8}\\y=-5\\z=\dfrac{1}{8}\end{matrix}\right.\)