Bài 22 :
\(a,\) Thay \(x=9\) vào \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{3+2}{3-5}=-\dfrac{5}{2}\)
\(b,B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\left(dkxd:x\ge0,x\ne25\right)\)
\(=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{1}{\sqrt{x}-5}\left(dpcm\right)\)
\(c,A=B.\left|x-4\right|\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-5}=\dfrac{1}{\sqrt{x}-5}.\left|x-4\right|\)
\(\Leftrightarrow\left|x-4\right|=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}.\left(\sqrt{x}-5\right)\)
\(\Leftrightarrow\left|x-4\right|=\sqrt{x}+2\)
\(\left|x-4\right|=\left\{{}\begin{matrix}x-4khix\ge4\\-x+4khix< 4\end{matrix}\right.\)
Với \(x\ge4\Rightarrow x-4=\sqrt{x}+2\Rightarrow x=9\left(tm\right)\)
Với \(x< 4\Rightarrow-x+4=\sqrt{x}+2\Rightarrow x=1\left(tm\right)\)
Vậy \(x=9,x=1\) thì \(A=B.\left|x-4\right|\)
Bài 23 :
\(a,\) Thay \(x=25\) vào A \(\Rightarrow A=\dfrac{7}{\sqrt{25}+8}=\dfrac{7}{5+8}=\dfrac{7}{13}\)
\(b,\) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\left(dkxd:x\ge0,x\ne9\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+8\sqrt{x}-3\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\left(dpcm\right)\)
\(c,\) \(P=A.B=\dfrac{7}{\sqrt{x}+8}.\dfrac{\sqrt{x}+8}{\sqrt{x}+3}=\dfrac{7}{\sqrt{x}+3}\)
\(\Rightarrow0< P\le\dfrac{7}{3}\)
Vậy các giá trị nguyên của P có thể đạt được là 1 và 2
Với \(P=1\), ta có : \(\dfrac{7}{\sqrt{x}+3}=1\Rightarrow\sqrt{x}+3=7\Rightarrow x=16\left(tm\right)\)
Với \(P=2\), ta có : \(\dfrac{7}{\sqrt{x}+3}=2\Rightarrow\sqrt{x}+3=\dfrac{7}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(P=A.B\) có giá trị số nguyên khi \(x=16\) hoặc \(x=\dfrac{1}{4}\)
