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ĐKXĐ: \(x\ge0;x\ne4;9\)
\(P=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}\right)=\frac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{x-4}\)
\(\frac{1}{P}\le-\frac{5}{2}\Leftrightarrow\frac{1}{P}+\frac{5}{2}\le0\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\)
\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\)
\(\Rightarrow\sqrt{x}\le\frac{1}{2}\Rightarrow0\le x\le\frac{1}{4}\)
Ta có \(a,\sqrt{9(x-1)}=21 \)
<=> \(3\sqrt{x-1}=21 \)
<=> \(\sqrt{x-1}=7 \)
<=>\(x-1=49\)
<=>x=50
b, \(\sqrt{4(x-1)^2}-6=0 \)
<=>\(2|x-1|-6=0\)
<=>\(|x-1|=3\)
<=>x=4 hoặc x=-2
c,\(\sqrt{(x-5)^2}=8 \)
<=>|x-5|=8
<=>x=-3 hoặc x=13
d,\(\sqrt{(2x-1)^2}=3 \)
<=>|2x-1|=3
=> x=2 hoặc x=-1
e, \(\sqrt{(2x+3)^2}=3 \)
<=>|2x+3|=3
=>x=0 hoặc x=-3
f, \(\sqrt{x^2-4x+4}=2x-3 \)
<=>\(\sqrt{(x-2)^2}=2x-3 \)
<=>|x-2|=2x-3
Với x-2=2x-3
=>x-1=0
<=>x=1
Với 2-x=2x-3
=>x=\(\frac{5}{3}\)