a: \(\left(x-1,5\right)^2=9\)
=>\(\left[{}\begin{matrix}x-1,5=3\\x-1,5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5+3=4,5\\x=-3+1,5=-1,5\end{matrix}\right.\)
b: \(\left(x+2\right)^3=-125\)
=>\(\left(x+2\right)^3=\left(-5\right)^3\)
=>x+2=-5
=>x=-5-2=-7
c: \(\left(\dfrac{2}{33}\right)^x\cdot11^x=\dfrac{4}{9}\)
=>\(\left(\dfrac{2}{33}\cdot11\right)^x=\dfrac{4}{9}\)
=>\(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
=>x=2
d: \(\left(x-2\right)^5=\left(x-2\right)^6\)
=>\(\left(x-2\right)^6-\left(x-2\right)^5=0\)
=>\(\left(x-2\right)^5\left[\left(x-2\right)-1\right]=0\)
=>\(\left(x-2\right)^5\cdot\left(x-3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
e: \(\dfrac{1}{4}\cdot2^x+2\cdot2^x=9\cdot2^6\)
=>\(\dfrac{9}{2}\cdot2^x=9\cdot2^6\)
=>\(2^{x-1}=2^6\)
=>x-1=6
=>x=7
f: \(\left(x+1,5\right)^4>=0\forall x;\left(y-3,5\right)^8>=0\forall y\)
Do đó: \(\left(x+1,5\right)^4+\left(y-3,5\right)^8>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+1,5=0\\y-3,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=3,5\end{matrix}\right.\)
