Bài 10:
\(A_{10}=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(-5\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A_{10}+\frac52=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+\frac52\)
\(=\frac{-10\sqrt{x}+4+5\sqrt{x}+15}{2\left(\sqrt{x}+3\right)}=\frac{-5\sqrt{x}+19}{2\left(\sqrt{x}+3\right)}\)
\(A_{10}<-\frac52\)
=>\(-5\sqrt{x}+19<0\)
=>\(-5\sqrt{x}<-19\)
=>\(\sqrt{x}>\frac{19}{5}\)
=>x>361/25
\(A_{10}>-\frac52\)
=>\(-5\sqrt{x}+19>0\)
=>\(-5\sqrt{x}>-19\)
=>\(\sqrt{x}<\frac{19}{5}\)
=>x<361/25
Kết hợp ĐKXĐ, ta được: 0<=x<361/25 và x<>1
\(A_{10}=-\frac52\)
=>\(-5\sqrt{x}+19=0\)
=>\(-5\sqrt{x}=-19\)
=>\(\sqrt{x}=\frac{19}{5}\)
=>\(x=\frac{361}{25}\) (nhận)
Bài 9:
\(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\frac{2-2\sqrt{x}}{3-\sqrt{x}}-1\)
\(=\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\)
\(=\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Ta có: \(A_9=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2-2\sqrt{x}}{3-\sqrt{x}}-1\right)\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{-3}{\sqrt{x}+3}\)
\(A_9<-\frac13\)
=>\(A_9+\frac13<0\)
=>\(\frac{-3}{\sqrt{x}+3}+\frac13<0\)
=>\(\frac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}<0\)
=>\(\sqrt{x}-6<0\)
=>\(\sqrt{x}<6\)
=>0<=x<36
Kết hợp ĐKXĐ, ta được; 0<=x<36 và x<>9
Bài 7:
\(A_7=\left(\frac{2\sqrt{x}-3}{x-1}+\frac{3\sqrt{x}}{x-\sqrt{x}}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\frac{2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3}{\sqrt{x}-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}-3+3\left(\sqrt{x}+1\right)}{\left.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right.}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{5\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{5}{\sqrt{x}-1}\)
\(A_7<-1\)
=>\(A_7+1<0\)
=>\(\frac{5}{\sqrt{x}-1}+1<0\)
=>\(\frac{5+\sqrt{x}-1}{\sqrt{x}-1}<0\)
=>\(\frac{\sqrt{x}+4}{\sqrt{x}-1}<0\)
=>\(\sqrt{x}-1<0\)
=>\(\sqrt{x}<1\)
=>0<=x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
