Bài 26 :
a) \(A=\dfrac{x+4}{6-3\sqrt{x}}\)
Khi \(x=9\Rightarrow A=\Leftrightarrow\dfrac{9+4}{6-3\sqrt{9}}=-\dfrac{13}{3}\)
b) \(A.B=\left(\dfrac{x+4}{6-3\sqrt{x}}\right).\left(\dfrac{5\sqrt{x}-2}{x-4}+\dfrac{2}{2-\sqrt{x}}\right)\)
\(=\dfrac{x+4}{-3\left(\sqrt{x}-2\right)}.\dfrac{5\sqrt{x}-2-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x+2}\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4}{-3\left(\sqrt{x}-2\right)}.\dfrac{5\sqrt{x}-2-2\sqrt{x}-4}{\left(\sqrt{x+2}\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4}{-3\left(\sqrt{x}-2\right)}.\dfrac{3\sqrt{x}-6}{\left(\sqrt{x+2}\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4}{-3\left(\sqrt{x}-2\right)}.\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x+2}\right)\left(\sqrt{x}-2\right)}=-\dfrac{x+4}{x-4}\)
Bài 25:
a: Thay x=16 vào A, ta được:
\(A=\dfrac{4-1}{4+2}=\dfrac{3}{6}=\dfrac{1}{2}\)
b: \(B=\dfrac{\sqrt[]{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)
\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+6}{\sqrt[]{x}-1}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+6}{\sqrt{x}+2}\)
Để P là số nguyên thì \(\sqrt{x}+6⋮\sqrt{x}+2\)
=>\(\sqrt{x}+2+4⋮\sqrt{x}+2\)
=>\(4⋮\sqrt{x}+2\)
mà \(\sqrt{x}+2>=2\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+2\in\left\{2;4\right\}\)
=>\(\sqrt{x}\in\left\{0;2\right\}\)
=>\(x\in\left\{0;4\right\}\)
Bài 27:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b: \(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
c: Thay x=4 vào P, ta được:
\(P=\dfrac{2+1}{2\cdot2}=\dfrac{3}{4}\)
