\(1,x=16\Rightarrow A=\dfrac{16-1}{\sqrt{16}}=\dfrac{15}{4}\)

\(2,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\left(dl:x>0,x\ne1\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{x-1}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{x-1}\\ =\dfrac{4x\sqrt{x}}{x-1}\)

\(3,P=A.B=\dfrac{x-1}{\sqrt{x}}.\dfrac{4x\sqrt{x}}{x-1}=4x\)

\(\sqrt{P}>P\Leftrightarrow\sqrt{4x}>4x\Leftrightarrow\left(\sqrt{4x}\right)^2>\left(4x\right)^2\Leftrightarrow4x>16x^2\Leftrightarrow4x-16x^2>0\Leftrightarrow4x\left(1-4x\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x>0\\1-4x>0\end{matrix}\right.\)

 \(\Leftrightarrow\left[{}\begin{matrix}x>0\\x< \dfrac{1}{4}\end{matrix}\right.\) 

Vậy \(S=\left\{x|0< x< \dfrac{1}{4}\right\}\) thì \(\sqrt{P}>P\)

\(4,\left|P\right|>P\Leftrightarrow\left|4x\right|>4x\)

\(TH_1:x\ge0\\4x>4x\Leftrightarrow4x-4x>0\Leftrightarrow0>0\left(VL\right) \)

\(TH_2:x< 0\\ -4x>4x\Leftrightarrow-4x-4x>0\Leftrightarrow-8x>0\Leftrightarrow x< 0\)

Vậy \(x< 0\) thì \(\left|P\right|>P\)

a) Từ trái qua phải 

\(Na_2O+H_2O\rightarrow2NaOH\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)

\(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\downarrow\)

b) Từ trái qua phải 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(Fe\left(OH\right)_2\xrightarrow[]{t^o}FeO+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(2FeCl_2+4H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+4HCl+SO_2\uparrow+2H_2O\)

D

D

chịu

3^x x 2 + 15 = 33

<=> 3^x = 9

<=> x = 2

3^x . 2 + 15 = 33

3^x . 2 = 33 - 15 = 18

3^x = 18 : 2 = 9 = 3²

=> x = 2

\(3^x.2+15=33\)

\(< =>3^x.2=33-15=18\)

\(< =>3^x=18:2=9\)

\(< =>x=2\)

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