1.
9x+1=73
9x=73-1
9x=72
x=72:9
x=8
Vậy x=8
2.
(x+2)(x-3)=0
⇔\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
Vậy x∈{-2;3}
3.
(2x-5)^2=9
(2x-5)^2=(±3)²
\(\left[ \begin{array}{l}2x-5=-3\\2x-5=3\end{array} \right.\)
\(\left[ \begin{array}{l}2x=2\\2x=8\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=4\end{array} \right.\)
Vậy x∈{1;4}
4.
Ta có :9=1.9=9.1=3.3=-1.(-9)=(-9).(-1)=-3.(-3)
mà x<y
=>x=1 thì y=9
x=-9 thì y=-1
Vậy (x;y)∈{(1;9);(-9;-1)}
