Question

Answer 1

a: \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)

\(=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

b: Thay x=9 vào P, ta được:

\(P=\frac{2\cdot3-1}{3+1}=\frac{6-1}{4}=\frac54\)

c: \(P=\frac12\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac12\)

=>\(4\sqrt{x}-2=\sqrt{x}+1\)

=>\(3\sqrt{x}=3\)

=>\(\sqrt{x}=1\)

=>x=1(loại)

d: P<1

=>P-1<0

=>\(\frac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}<0\)

=>\(\frac{\sqrt{x}-2}{\sqrt{x}+1}<0\)

=>\(\sqrt{x}-2<0\)

=>\(\sqrt{x}<2\)

=>0<=x<4

Kết hợp ĐKXĐ, ta được: 0<=x<4 và x<>1

e: \(P-2=\frac{2\sqrt{x}-1}{\sqrt{x}+1}-2=\frac{2\sqrt{x}-1-2\sqrt{x}-2}{\sqrt{x}+1}=\frac{-3}{\sqrt{x}+1}<0\)

=>P<2∀x thỏa mãn ĐKXĐ