Question

1. Với x >= 0 , cho biểu thức P = (1/(sqrt(x) - 1) - (2sqrt(x))/(x * sqrt(x) - x + sqrt(x) - 1))((x + sqrt(x))/(x * sqrt(x) + x + sqrt(x) + 1) + 1/(x + 1))
a. Rút gọn P
b. Tìm giá trị của x để P < 1/2
c. Tìm giá trị của x để P = 1/3
d. Tìm x nguyên để P nguyên

Answer 1

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\cdot\sqrt{x}-x+\sqrt{x}-1}\)

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

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

Ta có: \(\frac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\)

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

\(=\frac{\sqrt{x}}{x+1}+\frac{1}{x+1}=\frac{\sqrt{x}+1}{x+1}\)

Ta có: \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\cdot\sqrt{x}-x+\sqrt{x}-1}\right)\cdot\left(\frac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right)\)

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

b: \(P<\frac12\)

=>\(P-\frac12<0\)

=>\(\frac{x-1}{\left(x+1\right)^2}-\frac12<0\)

=>\(\frac{2x-2-\left(x+1\right)^2}{2\left(x+1\right)^2}<0\)

=>\(2x-2-\left(x+1\right)^2<0\)

=>\(2x-2-x^2-2x-1<0\)

=>\(-x^2-3<0\)

=>\(x^2+3>0\) (luôn đúng với mọi x thỏa mãn ĐKXĐ)

vậy: \(\begin{cases}x\ge0\\ x<>1\end{cases}\)

c: \(P=\frac13\)

=>\(\frac{x-1}{\left(x+1\right)^2}=\frac13\)

=>\(\left(x+1\right)^2=3\left(x-1\right)\)

=>\(x^2+2x+1=3x-3\)

=>\(x^2-x+4=0\)

=>\(x^2-x+\frac14+\frac{15}{4}=0\)

=>\(\left(x-\frac12\right)^2+\frac{15}{4}=0\) (vô lý)