Bài 5:
a: \(B=\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{x-9}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\right.\sqrt{x}-3)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)
b: \(B=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}+3+5}{\sqrt{x}+3}=1+\frac{5}{\sqrt{x}+3}\le1+\frac53=\frac83\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
c: \(P=A\cdot B\)
\(=\frac{7}{\sqrt{x}+8}\cdot\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)
Để P nguyên thì 7⋮\(\sqrt{x}+3\)
=>\(\sqrt{x}+3=7\)
=>\(\sqrt{x}=4\)
=>x=16(nhận)
Bài 2:
a: \(P=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\frac{x-1}{\sqrt{x}}:\left(\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1-\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: P=9
=>\(\left(\sqrt{x}+1\right)^2=9\sqrt{x}\)
=>\(x-7\sqrt{x}+1=0\)
=>\(x-7\sqrt{x}+\frac{49}{4}-\frac{45}{4}=0\)
=>\(\left(\sqrt{x}-\frac72\right)^2=\frac{45}{4}\)
=>\(\sqrt{x}-\frac72=\frac{3\sqrt5}{2}\)
=>\(\sqrt{x}=\frac{3\sqrt5+7}{2}\)
=>\(x=\frac{45+49+2\cdot3\sqrt5\cdot7}{4}=\frac{94+42\sqrt5}{4}=\frac{47+21\sqrt5}{2}\) (nhận)
