"Em" thường dùng cho h/s tiểu học và THCS ; xưng hô nhẹ nhàng ; yêu quý . Còn anh/chị thường dùng cho h/s THPT ; xưng hô thể hiện sự tôn trọng với những người đã lớn ; có sự hiểu biết 

ĐK : \(x\ge0\)

P/t \(\Leftrightarrow x\left(x^2+2\sqrt{x}.x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+2x\sqrt{x}+2=0\left(PTVN;x\ge0\right)\end{matrix}\right.\)

\(\dfrac{2001}{2003}=1-\dfrac{2}{2003};\dfrac{2004}{2006}=1-\dfrac{2}{2006}\)

Thấy : \(\dfrac{2}{2003}>\dfrac{2}{2006}\Rightarrow1-\dfrac{2}{2003}< 1-\dfrac{2}{2006}\) 

\(\Rightarrow\dfrac{2001}{2003}< \dfrac{2004}{2006}\)

\(\dfrac{6}{7}=1-\dfrac{1}{7};\dfrac{100}{102}=1-\dfrac{2}{102}=1-\dfrac{1}{51}\)

Mà : \(\dfrac{1}{7}>\dfrac{1}{51}\Rightarrow1-\dfrac{1}{7}< 1-\dfrac{1}{51}\)  \(\Rightarrow\dfrac{6}{7}< \dfrac{100}{102}\)

\(-\dfrac{1}{4}.\dfrac{152}{11}+\dfrac{68}{4}.\dfrac{-1}{11}=-\dfrac{1}{11}\left(\dfrac{68}{4}+\dfrac{152}{4}\right)=-\dfrac{1}{11}.\dfrac{220}{4}=-5\)

\(n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\)

Ta có : \(\left\{{}\begin{matrix}27a+56b=11\\3a+2b=\dfrac{8,96}{22,4}.2\left(BTe\right)\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,2.27}{11}.100\%=49,09\%\) ; \(\%m_{Fe}=50,91\%\)

Có : \(n_{SO_4^{2-}}=n_{H_2}=0,4\left(mol\right)\)

\(m_{muoi}=m_{KL}+m_{SO_4^{2-}}=11+0,4.96=49,4\left(g\right)\)

a) \(\sqrt{7}-\sqrt{6}=\dfrac{1}{\sqrt{7}+\sqrt{6}}< \dfrac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\)  ( đpcm ) 

b) \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)  = \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)  

\(=2+\sqrt{3}-\left(2-\sqrt{3}\right)=2\sqrt{3}\)

c) ĐKXĐ : x > 0 

Ta có : \(A=\dfrac{x+1+\sqrt{x}}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge2+1=3\)  ( BĐT Cô - si ) 

' = ' \(\Leftrightarrow x=1\)

ĐKXĐ : x \(\ge0\)\(;B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)  . " = " \(\Leftrightarrow x=0\)

d)  ĐKXĐ : \(-5\le x\le5\)

\(A=\sqrt{25-x^2}\ge0\)  . " = " \(\Leftrightarrow x=\pm5\)

\(A\le5\) . " = " \(\Leftrightarrow x=0\)

ĐKXĐ : \(2-\sqrt{13}\le x\le2+\sqrt{13}\)

\(B=\sqrt{9+4x-x^2}=\sqrt{13-\left(x^2-4x+4\right)}=\sqrt{13-\left(x-2\right)^2}\le\sqrt{13}\) \(\forall x\)

" = " \(\Leftrightarrow x=2\)  . Max B = \(\sqrt{13}\Leftrightarrow x=2\)

\(B=\sqrt{9+4x-x^2}\ge0\) . " = " \(\Leftrightarrow9+4x-x^2=0\Leftrightarrow x=2\pm\sqrt{13}\)

Min B = \(0\Leftrightarrow x=2\pm\sqrt{13}\)