Question

Cho các biểu thức \( A = \frac{\sqrt{x} - 1}{\sqrt{x} - 3} \) và \( B = \frac{2\sqrt{x}}{\sqrt{x} + 3} - \frac{\sqrt{x}}{3 - \sqrt{x}} \) với \( x \geq 0; x \neq 9 \).

a) Tính giá trị của \( A \) khi \( x = 16 \).

b) Rút gọn biểu thức \( P \) biết \( P = B : A \).

c) Tìm \( x \) sao cho \( P = \frac{1}{2} \).

d) Tìm \( x \) để \( P = P^2 \).

Answer 1

a: Khi x=16 thì \(A=\dfrac{4-1}{4-3}=\dfrac{3}{1}=3\)

b: P=B:A

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

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

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{3x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

c: \(P=\dfrac{1}{2}\)

=>\(\dfrac{3\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\)

=>\(6\sqrt{x}=\sqrt{x}+3\)

=>\(5\sqrt{x}=3\)

=>\(\sqrt{x}=\dfrac{3}{5}\)

=>\(x=\dfrac{9}{25}\left(nhận\right)\)

d: \(P^2=P\)

=>\(P^2-P=0\)

=>P(P-1)=0

=>\(\left[{}\begin{matrix}P=0\\P=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3\sqrt{x}}{\sqrt{x}+3}=0\\\dfrac{3\sqrt{x}}{\sqrt{x}+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\3\sqrt{x}=\sqrt{x}+3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{9}{4}\left(nhận\right)\end{matrix}\right.\)