1: Sửa đề: \(A=\dfrac{4\sqrt{x}+1}{\sqrt{x}-4}\)
Thay x=9 vào A, ta được:
\(A=\dfrac{4\cdot3+1}{3-4}=\dfrac{12+1}{-1}=-13\)
b: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{x-16}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\)
\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)+5\sqrt{x}+12}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\)
\(=\dfrac{x-\sqrt{x}-12+5\sqrt{x}+12}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\)
\(=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-4}\)
1. Thay x = 9 vào A ta có:
\(A=\dfrac{4\cdot\sqrt{9+1}}{\sqrt{9}-4}=\dfrac{4\sqrt{10}}{3-4}=\dfrac{4\sqrt{10}}{-1}=-4\sqrt{10}\)
2. \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{x-16}\)
\(B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}+\dfrac{5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{x+3\sqrt{x}-4\sqrt{x}-12+5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-4}\)
3. Ta có: \(P=A:B\)
\(P=\dfrac{4\sqrt{x+1}}{\sqrt{x}-4}:\dfrac{\sqrt{x}}{\sqrt{x}-4}\)
\(P=\dfrac{4\sqrt{x+1}}{\sqrt{x}-4}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}}\)
\(P=\dfrac{4\sqrt{x+1}}{\sqrt{x}}\)
Mà: \(P\sqrt{x}+3=8+x+\dfrac{1}{4}\sqrt{x-3}\left(x\ge3\right)\)
\(\Leftrightarrow\dfrac{4\sqrt{x+1}}{\sqrt{x}}\cdot\sqrt{x}+3=8+x+\dfrac{1}{4}\sqrt{x-3}\)
\(\Leftrightarrow4\sqrt{x+1}=5+x+\dfrac{1}{4}\sqrt{x-3}\)
\(\Leftrightarrow\left(x+1\right)-4\sqrt{x+1}+4=-\dfrac{1}{4}\sqrt{x-3}\)
\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2=-\dfrac{1}{4}\sqrt{x-3}\)
Ta có:
\(\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2\ge0\forall x\\-\dfrac{1}{4}\sqrt{x-3}\le0\forall x\end{matrix}\right.\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\sqrt{x+1}-2=0\\-\dfrac{1}{4}\sqrt{x-3}=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(tm\right)\)
