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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 \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)
thực hiện phép tính bằng -4,8172906
còn lại dễ bạn tự làm nhé!!!
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 \(\left|2x-27\right|^{2011}=\left(3y+10\right)^{2012}=0\Leftrightarrow\hept{\begin{cases}\left|2x-27\right|=0\\\left(3y+10\right)^{2012}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)
\(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2018}=0\)
Ta có \(\left|2x-27\right|^{2017}\ge0\forall x;\left(3y+10\right)^{2018}\ge0\forall y\)
\(\Rightarrow\left|2x-27\right|^{2017}+\left(3.y+10\right)^{2018}\ge0\forall x;y\)
\(\Rightarrow\left|2x-17\right|^{2017}+\left(3y+10\right)^{2018}=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-17=0\\3.y+10=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{17}{2}\\y=-\frac{10}{3}\end{cases}}\)
|2x-27|^2011>0
(3y+10)^2>0
=|2x-27|^2011+(3y+10)^2>0
mà |2x-27|^2011+(3y+10)^2=0
=>|2x-27|^2011=(3y+10)^2=0
+)|2x-27|^2011=0=>2x-27=0=>2x=27=>x=13,5
+)(3y+10)^2=0=>3y+10=0=>3y=-10=>y=-10/3
Ta có :
\(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)
Mà \(\left\{{}\begin{matrix}\left|2x-27\right|^{2007}\ge0\\\left(3y+10\right)^{2008}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|2x-27\right|^{2007}=0\\\left(3y+10\right)^{2008}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy ..