Question

P= \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\)

a) Tìm điều kiện xác định và rút gọn P

b) Tìm x để P=\(\dfrac{2}{7}\)

c) Cho M=P.\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-2}\). Tìm x nguyên để M nguyên

d) So sánh P và \(\dfrac{1}{3}\) với mọi x thỏa mãn đkxđ

Answer 1

\(a.P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\) ( x # 1 ; x ≥ 0 )

\(b.P=\dfrac{2}{7}\) ⇔ \(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

⇔ \(7\sqrt{x}-2x-2\sqrt{x}-2=0\text{⇔}-2x+5\sqrt{x}-2=0\text{⇔}\left(2\sqrt{x}-1\right)\left(2-\sqrt{x}\right)=0\) ⇔\(x=\dfrac{1}{4}\left(TM\right)orx=4\left(TM\right)\)

\(c.M=P.\dfrac{x+\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\)

Để : M ∈ Z ⇔ \(\dfrac{2}{\sqrt{x}-2}\) ∈ Z ⇔ \(\sqrt{x}-2\text{∈ }\left\{1;-1;2;-2\right\}\)

+) \(\sqrt{x}-2=1\text{⇔}x=9\left(TM\right)\)

+) \(\sqrt{x}-2=-1\text{⇔}x=1\left(KTM\right)\)

+) \(\sqrt{x}-2=2\text{⇔}x=16\left(TM\right)\)

+) \(\sqrt{x}-2=-2\text{⇔}x=0\left(TM\right)\)

\(d.P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\text{≥}1>\dfrac{1}{3}\left(x\text{≥}0\right)\)