a) \(3\sqrt{2}-2\sqrt{3}=\sqrt{3}.\sqrt{3}.\sqrt{2}-\sqrt{2}.\sqrt{2}.\sqrt{3}=\left(\sqrt{3}-\sqrt{2}\right).\sqrt{6}\)

b) \(\sqrt{2}+\sqrt{6}+\sqrt{14}+\sqrt{42}=\left(\sqrt{3}+1\right)\sqrt{2}+\sqrt{14}\left(\sqrt{3}+1\right)=\sqrt{2}\left(\sqrt{7}+1\right)\left(\sqrt{3}+1\right)\)

c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{2}}=\sqrt{\dfrac{9}{4}}\)

a) Gọi n_FeS = a (mol)

       \(n_{FeS_2}=b\left(mol\right)\) với a; b > 0

\(n_{SO_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Ta có: \(\left\{{}\begin{matrix}m_{hh}=17,8=m_{FeS}+m_{FeS_2}=88a+120b\\n_{S\left(SO_2\right)}=0,25=n_{FeS}+2n_{FeS_2}\left(bt\left[S\right]\right)=a+2b\end{matrix}\right.\)

=> a = 0,1(mol); b = 0,075(mol)

m_FeS= n.M= 0,1 . 88 = 8,8(g)

=> \(m_{FeS_2}=m_{hh}-m_{FeS}=17,8-8,8=9\left(g\right)\)

b) PT:

\(4FeS+7O_2\underrightarrow{t^o}2Fe_2O_3+4SO_2\uparrow\\ 4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow\)

\(Theo2pt\Rightarrow n_{O_2}=\dfrac{7n_{FeS}+11n_{FeS_2}}{4}=0,38125\left(mol\right)\)

\(\Rightarrow V_{O_2}=n\cdot22,4=0,38125\cdot22,4=8,54\left(l\right)\)

1. patient

2. generous

3. shy

4. kind

5. lazy

6. pessimistic

%m chất tan ở đây e tưởng phải tính theo m_dd chứ ạ

a) M \(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\right)\)

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

        = \(\dfrac{x}{\sqrt{x}-1}\)

b) Ta có \(\dfrac{x}{\sqrt{x}-1}\)= \(\dfrac{-1}{2}\)

\(\Leftrightarrow2x=1-\sqrt{x}\)

\(\Leftrightarrow2x+\sqrt{x}-1=0\)

\(\Leftrightarrow2x+2\sqrt{x}-\sqrt{x}-1\) \(=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

Vì:

 \(\sqrt{x}+1>0\Rightarrow2\sqrt{x}-1=0\\ \Rightarrow x=\dfrac{1}{4}\left(tm\right)\)

S_xq= 2 \(\pi.r.h\)

\(\Leftrightarrow352\simeq2\cdot3,14\cdot7\cdot h\)

\(\Rightarrow\) h = \(8\left(cm\right)\)