g)$(x-1)^{x+2}$ =$(x-1)^{x+6}$
⇒$(x-1)^{x}$.$(x-1)^{2}$ =$(x-1)^{x}$.$(x-1)^{6}$
⇒$(x-1)^{2}$ =$(x-1)^{6}$
⇒$(x-1)^{6}$-$(x-1)^{2}$=0
⇒$(x-1)^{2}$($(x-1)^{4}$$-1^{}$ )=0
⇒\(\left[ \begin{array}{l}(x-1)^{2}=0\\(x-1)^{4}=1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x-1^{}=0\\x-1^{}=1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x^{}=1\\x^{}=2\end{array} \right.\)
Vậy$x^{}$ ∈{$1^{} ;2$}
h)$(x+20)^{100}$ +|$y^{}+4$|=$0^{}$
vì $\left \{ {{(x+20)^{100}≥0} \atop {|y^{}+4|≥0}} \right.$
mà$(x+20)^{100}$ +|$y^{}+4$|=$0^{}$
⇒$\left \{ {{(x+20)^{100}=0} \atop {|y^{}+4|=0}} \right.$
⇒$\left \{ {{x+20^{}=0} \atop {y^{}+4=0}} \right.$
⇒$\left \{ {{x^{}=-20} \atop {y^{}=-4}} \right.$
I)$2,5x^{}$+$\frac{3}{4}$ =$3,5x^{}$
⇔$3,5x^{}$$-2,5x^{}$ = $\frac{3}{4}$
⇒$x^{}$ =$\frac{3}{4}$
Câu k mk ko hiểu đề lắm
-Mời bạn tham khảo !
