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Vì \(\left|2x-27\right|\ge0\Rightarrow\left|2x-27\right|^{2011}\ge0\); \(\left(3y+10\right)^{2012}\ge0\)
=>\(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)
Dấu "=" xảy ra khi \(\left|2x-27\right|^{2011}=\left(3y+10\right)^{2012}=0\Leftrightarrow\hept{\begin{cases}\left|2x-27\right|=0\\\left(3y+10\right)^{2012}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)
1,
Vì \(\left|2x-27\right|^{2007}\ge0;\left(3y+10\right)^{2008}\ge0\)
\(\Rightarrow\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\)
Mà \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)
\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2007}=0\\\left(3y+10\right)^{2008}=0\end{cases}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-10}{3}\end{cases}}}\)
2,
TH1: \(x\ge\frac{3}{5}\)
<=> 2(5x-3)-2x=14
<=> 10x-6-2x=14
<=>8x-6=14
<=>8x=20
<=>x=5/2 (thỏa mãn)
TH2: x < 3/5
<=> 2(3-5x)-2x=14
<=>6-10x-2x=14
<=>6-12x=14
<=>12x=-8
<=>x=-2/3 (thỏa mãn)
Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)
Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\left(\forall x\right)\\\left(3y+4\right)^{2020}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\left(\forall x,y\right)\)
Mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\left(\forall x,y\right)\)
\(\Rightarrow\hept{\begin{cases}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)
Khi đó thay vào ta được:
\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)
\(\Rightarrow M=-\frac{1159}{36}\)
\(\left|2x-27\right|^{2017}+\left(3y+27\right)^{2016}=0\)
\(\Rightarrow\left|2x-27\right|^{2017}=0\) và \(\left(3y+27\right)^{2016}=0\)
+) \(\left|2x-27\right|^{2017}=0\Rightarrow2x-27=0\Rightarrow2x=27\Rightarrow x=\frac{27}{2}\)
+) \(\left(3y+27\right)^{2016}=0\Rightarrow3y+27=0\Rightarrow3y=-27\Rightarrow y=-9\)
Vậy \(x=\frac{27}{2};y=-9\)
ta có:
|2x-27|2017≥0
(3y+27)2016 ≥0
vậy |2x-27|2017+(3y+37)2016 ≥0
dấu "=" xảy ra khi
|2x-27|2017=(3y+27)2016=0
|2x-27|2017=0
=> 2x=27
=>x=27/2
(3y+27)2016=0
=> 3y=-27
=> y=-9
vậy với x=27/2 và y=-9 thì x,y thỏa mãn yêu cầu đề bài