\(B1:a;M=\dfrac{A}{B}=\dfrac{\dfrac{x+12}{\sqrt{x}-1}}{\dfrac{\sqrt{x}+2}{\sqrt{x}-1}}=\dfrac{x+12}{\sqrt{x}+2}=4+\dfrac{x-4\sqrt{x}+4}{\sqrt{x}+2}=4+\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+2}\ge4\Rightarrow min=4\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow x=4\left(tm\right)\)
\(b,\dfrac{A}{B}=\dfrac{\left(x+2\sqrt{x}\right)\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x}{\sqrt{x}-1}>1\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(do:x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
\(3;M=\dfrac{B}{A}=\dfrac{x-1}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
\(\Rightarrow\dfrac{1}{m}-\dfrac{\sqrt{x}+1}{8}\ge1\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)
\(\Leftrightarrow\dfrac{16\sqrt{x}-\left(\sqrt{x}+1\right)^2-8\sqrt{x}-8}{8\sqrt{x}+8}\ge0\Rightarrow16\sqrt{x}-x-2\sqrt{x}-1-8\sqrt{x}-8\ge0\Leftrightarrow6\sqrt{x}-x-9\ge0\Leftrightarrow-\left(x-6\sqrt{x}+9\right)\ge0\Leftrightarrow\left(\sqrt{x}-3\right)^2\le0\Leftrightarrow x=9\left(tm\right)\)
\(\)\(d;đk:x\ge0;x\ne4\Rightarrow P=A.B=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(\Rightarrow\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\right|>\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(TH1:\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\ge0\Leftrightarrow\sqrt{x}-1\ge0\Leftrightarrow x\ge1\left(x\ne4\right)\Rightarrow\)
\(P>P\left(vô-lý\right)\)
\(TH2:P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow0\le x< 1\)
\(\Rightarrow-P>P\Leftrightarrow-2P>0\Leftrightarrow P< 0\Leftrightarrow0\le x< 1\)
giúp e với ạ e cảm ơnn
