Giải câu 3 trang 33 sách phát triển năng lực toán 9 tập 1.
a, Với x > 0 ta có:
P = $(\frac{2\sqrt{x}+1}{x+\sqrt{x}}-\frac{1-\sqrt{x}}{\sqrt{x}}):(1+\frac{2}{\sqrt{x}})$
= $(\frac{2\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}+1)}-\frac{1-\sqrt{x}}{\sqrt{x}}):\frac{\sqrt{x}+2}{\sqrt{x}}$
= $(\frac{2\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}+1)}-\frac{(1-\sqrt{x}).(\sqrt{x}+1)}{\sqrt{x}.(\sqrt{x}+1)}).\frac{\sqrt{x}}{\sqrt{x}+2}$
= $\frac{2\sqrt{x}+1-(1-x)}{\sqrt{x}.(\sqrt{x}+1)}.\frac{\sqrt{x}}{\sqrt{x}+2}$
= $\frac{2\sqrt{x}+x}{\sqrt{x}+1}.\frac{1}{\sqrt{x}+2}$ = $\frac{\sqrt{x}.(\sqrt{x}+2)}{\sqrt{x}+1}.\frac{1}{\sqrt{x}+2}$ = $\frac{\sqrt{x}}{\sqrt{x}+1}$
b, $x = 2019 - 2\sqrt{2018}=2018-2\sqrt{2018}+1=(\sqrt{2018}-1)^{2}$
Thay $x=(\sqrt{2018}-1)^{2}$ vào P ta có:
P = $\frac{\sqrt{(\sqrt{2018}-1)^{2}}}{\sqrt{(\sqrt{2018}-1)^{2}}+1}$ = $\frac{\sqrt{2018}-1}{\sqrt{2018}-1+1}$ = $\frac{\sqrt{2018}-1}{\sqrt{2018}}$