a) ĐKXĐ: x$\neq$ ±1
b) P = ($\frac{3}{x-1}$ + $\frac{1}{\sqrt{x}+1}$ ) : $\frac{1}{\sqrt{x}+1}$
P = ($\frac{3}{(\sqrt{x}+1)(\sqrt{x}-1)}$ + $\frac{1}{\sqrt{x}+1}$ ) : $\frac{1}{\sqrt{x}+1}$
P = ($\frac{3}{(\sqrt{x}+1)(\sqrt{x}-1)}$ + $\frac{\sqrt{x}-1 }{(\sqrt{x}+1)(\sqrt{x}-1)}$ ) .$\sqrt{x}+1$
p = ($\frac{3 + \sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}$ ) . $\sqrt{x}+1$
P = $\frac{\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}-1)}$ . $\sqrt{x}+1$
P = $\frac{\sqrt{x}+2}{\sqrt{x}-1}$
Vậy P = $\frac{\sqrt{x}+2}{\sqrt{x}-1}$
c) Để P = $\frac{5}{4}$ <=> $\frac{\sqrt{x}+2}{\sqrt{x}-1}$ = $\frac{5}{4}$
<=>5($\sqrt{x}-1$) = 4($\sqrt{x}+2$)
<=>5$\sqrt{x}$ - 5 = 4$\sqrt{x}$ + 8
<=> 5$\sqrt{x}$ - 4 $\sqrt{x}$= 8 + 5
<=> $\sqrt{x}$ = 13
<=> x = 169 (tmđk)
Vậy x = 169 thì P = $\frac{5}{4}$.
Cho mk ctlhn và vote 5 sao cho mk nha!!!!
