1) thay x= 9
⇒A= $\frac{√9+2}{√9-5}$
⇒A= $\frac{3+2}{3-5}$
⇒A= $\frac{5}{-2}$
vậy A= $\frac{5}{-2}$ ⇔x=9
2)B= $\frac{3}{√x+5}$ + $\frac{20-2√x}{x-25}$
B= $\frac{3(√x-5)}{(√x+5)(√x-5)}$ + $\frac{20-2√x}{(√x+5)(√x-5)}$
B= $\frac{3√x-15+20-2√x)}{(√x+5)(√x-5)}$
B= $\frac{5+√x}{(√x+5)(√x-5)}$
B= $\frac{1}{√x-5}$(đccm)
3)để A=B.|x-4|
⇔$\frac{√x+2}{√x-5}$ = ($\frac{1}{√x-5}$).|x-4|
⇔$\frac{√x+2}{√x-5}$-($\frac{|x-4|}{√x-5}$)=0
⇔$\frac{(√x+2)-|x-4|}{√x-5}$=0
⇔(√x+2)-|x-4|=0
⇔√x+2=|x-4|
⇔\(\left[ \begin{array}{l}√x+2=x-4\\√x+2=4-x\end{array} \right.\)
⇔\(\left[ \begin{array}{l}√x+2=(√x-2)(√x+2)\\√x+2=(2-√x)(√x+2)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}√x-2=1\\2-√x=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}√x=3\\√x=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=9\\x=1\end{array} \right.\)
vậy để A=B.|x-4|⇔x∈{9;1}
