1: \(\left(\dfrac{3x-5}{9}\right)^{2018}>=0\forall x\)
\(\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall y\)
Do đó: \(\left(\dfrac{3x-5}{9}\right)^{2018}+\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall x,y\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\\\dfrac{3y+0,4}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\3y+0,4=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{0.4}{3}=\dfrac{-2}{15}\end{matrix}\right.\)
3: \(\left(x-\dfrac{1}{2}\right)^4>=0\forall x\)
\(\left(x+1\right)^4>=0\forall x\)
Do đó: \(\left(x-\dfrac{1}{2}\right)^4+\left(x+1\right)^4>=0\forall x\)
mà \(\left(x-\dfrac{1}{2}\right)^4+\left(x+1\right)^4< =0\forall x\)
nên \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+1=0\end{matrix}\right.\)
=>x=1/2 và x=-1
=>\(x\in\varnothing\)
5: \(\left(\dfrac{1}{2}-x\right)\cdot\left(1+x\right)>0\)
=>\(\left(x-\dfrac{1}{2}\right)\left(x+1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{1}{2}>0\\x+1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x< -1\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{1}{2}< 0\\x+1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x>-1\end{matrix}\right.\)
=>-1<x<1/2
