Tính A = tan 20o.tan 80o + tan 80o.tan 140o + tan 140o.tan 20o
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ở $20^oC$,
300 gam nước hòa tan 108,6 gam NaCl thu được dung dịch bão hòa
Suy ra :
100 gam nước hòa tan x gam NaCl thu được dung dịch bão hòa
$\Rightarrow x = \dfrac{100.108,6}{300} = 36,2(gam)$
$\Rightarrow S_{NaCl} = 36,2(gam)$
b)
$S_{AgNO_3} = 222(gam)$, tức là :
222 gam $AgNO_3$ tan tối đa trong 100 gam nước tạo thành dung dịch bão hòa
Suy ra:
y gam $AgNO_3$ tan tối đa trong 250 gam nước tạo thành dung dịch bão hòa
$\Rightarrow y = \dfrac{250.222}{100} = 2,50222(gam)$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,S_{KNO_3\left(20^oC\right)}=\dfrac{60}{190}.100=31,6\left(g\right)\)
\(b,m_{H_2O}=\dfrac{69,9}{39,8+100}.100=50\left(g\right)\\ \rightarrow m_{NaCl\left(tách,ra\right)}=\dfrac{50}{100}.\left(39,8-36\right)=1,9\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S_{KNO3\left(20oC\right)}=\frac{60}{190}.100=31,58\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 13: nNa= 0,2 mol ; nK= 0,1 mol
2Na + 2H2O → 2NaOH + H2↑
0,2 mol 0,2 mol 0,1 mol
2K + 2H2O → 2KOH + H2↑
0,1 mol 0,1 mol 0,05 mol
a) tổng số mol khí H2 là: nH2= 0,1 + 0,05 = 0,15 mol
→VH2= 0,15 x 22,4 = 3,36 (l)
b) mNaOH= 0,2 x 40= 8 (g) ; mKOH= 0,1 x 56= 5,6 (g)
mdung dịch= mNa + mK + mH2O - mH2 = 4,6 + 3,9 + 91,5 - 0,15x2 = 99,7 (g)
→C%NaOH= 8/99,7 x100%= 8,02%
→C%KOH= 5,6/99,7 x100%= 5,62%
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\begin{array}{l}2a = \left( {a + b} \right) + \left( {a - b} \right) \Rightarrow \tan 2a = \tan \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right]\\2b = \left( {a + b} \right) - \left( {a - b} \right) \Rightarrow \tan 2b = \tan \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right]\end{array}\)
\(\begin{array}{l}\tan \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right] = \frac{{\tan \left( {a + b} \right) + \tan \left( {a - b} \right)}}{{1 - \tan \left( {a + b} \right).\tan \left( {a - b} \right)}} = \frac{{3 + 2}}{{1 - 3.2}} = - 1\\\tan \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right] = \frac{{\tan \left( {a + b} \right) - \tan \left( {a - b} \right)}}{{1 + \tan \left( {a + b} \right).\tan \left( {a - b} \right)}} = \frac{{3 - 2}}{{1 + 3.2}} = \frac{1}{7}\end{array}\)
Vậy \(\tan 2a = - 1,\,\,\,\tan 2b = \frac{1}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\tan \left( {a + b} \right) = \frac{{\sin \left( {a + b} \right)}}{{\cos \left( {a + b} \right)}} = \frac{{\sin a.\cos b + \cos a.\sin b}}{{\cos a.\cos b - \sin a.\sin b}}\)
\(\begin{array}{l} = \frac{{\sin a.\cos b + \cos a.\cos b}}{{\cos a.\cos b - \sin a.\sin b}} = \frac{{\sin a.\cos b}}{{\cos a.\cos b - \sin a.\sin b}} + \frac{{\cos a.\sin b}}{{\cos a.\cos b - \sin a.\sin b}}\\ = \frac{{\frac{{\sin a.\cos b}}{{\cos a.\cos b}}}}{{\frac{{\cos a.\cos b - \sin a.\sin b}}{{\cos a.\cos b}}}} + \frac{{\frac{{\cos a.\sin b}}{{\cos a.\cos b}}}}{{\frac{{\cos a.\cos b - \sin a.\sin b}}{{\cos a.\cos b}}}} = \frac{{\tan a}}{{1 - \tan a.\tan b}} + \frac{{\tan b}}{{1 - \tan a.\tan b}}\\ = \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\end{array}\)
\( \Rightarrow \tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\)
b)
\(\tan \left( {a - b} \right) = \tan \left( {a + \left( { - b} \right)} \right) = \frac{{\tan a + \tan \left( { - b} \right)}}{{1 - \tan a.\tan \left( { - b} \right)}} = \frac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}\)
Bấm máy tính được không ta
\(A=\frac{sin80}{cos80}\left(\frac{sin20}{cos20}+\frac{sin140}{cos140}\right)+\frac{sin140.sin20}{cos140.cos20}\)
\(=\frac{sin80}{cos80}\left(\frac{sin20.cos140+cos20.sin140}{cos20.cos140}\right)+\frac{\frac{1}{2}\left(cos120-cos160\right)}{cos20.cos140}\)
\(=\frac{sin80}{cos80}.\frac{sin160}{cos20.cos140}+\frac{cos120-cos160}{2cos20.cos140}\)
\(=\frac{2sin^280}{cos20.cos140}+\frac{cos120-cos160}{2cos20.cos140}=\frac{1-cos160}{cos20.cos140}+\frac{cos120-cos160}{2cos20.cos140}\)
\(=\frac{2-2cos160+cos120-cos160}{2cos20.cos140}=\frac{\frac{3}{2}-3cos160}{cos120+cos160}=\frac{-3\left(-\frac{1}{2}+cos160\right)}{-\frac{1}{2}+cos160}=-3\)