$=x^2-2xy-5xy+10y^2\\=x(x-2y)-5y(x-2y)\\=(x-5y)(x-2y)$

$=\dfrac{\cot 26^o}{\cot 26^o}-1\\=1-1=0$

Vậy $E=0$

a/ $3\sqrt 7=\sqrt{63}$

$2\sqrt{15}=\sqrt{60}$

Ta có: 63>60

$\Rightarrow\sqrt{63}>\sqrt{60}$ hay $3\sqrt 7>2\sqrt{15}$

b/ $-4\sqrt 5=-\sqrt{80}$

$-5\sqrt 3=-\sqrt{75}$

Ta có: 80>75

$\Rightarrow \sqrt{80}>\sqrt{75}$

$\Rightarrow-\sqrt{80}<-\sqrt{75}$ hay $-4\sqrt 5<-5\sqrt 3$

$U_3=I_3.R_3=0,3.10=3V$

$U_3=U_2=3V$

$\to I_2=\dfrac{U_2}{R_2}=\dfrac{3}{15}=0,2A$

Vậy $I_2=0,2A$

$R_{td}=R_1+\dfrac{R_2R_3}{R_2+R_3}=30\Omega$

$I=\dfrac{U}{R_{td}}=\dfrac{60}{30}=2A$

$I_1=I_{23}=I=2A$

$U_{23}=I_{23}.R_{23}=2.\dfrac{30.20}{30+20}=24V$

$U_2=U_3=24V$

$I_2=\dfrac{U_2}{R_2}=\dfrac{24}{30}=0,8A$

$I_3=\dfrac{U_3}{R_3}=\dfrac{24}{20}=1,2A$

Vậy $I_1=2A;I_2=0,8A;I_3=1,2A$

a/ $d_{CO/N_2}=\dfrac{28}{28}=1$

b/ $d_{CO_2/O_2}=\dfrac{32}{32}=1$

c/ $d_{N_2/H_2}=\dfrac{28}{2}=14$

e/ $d_{H_2S/H_2}=\dfrac{34}{2}=17$

e/ $Na_2CO_3$

PTK=23.2+12+16.3=106 đvC

f/ $FeCl_3$

PTK=56+35,5.3=162,5 đvC

g/ $BaS$

PTK=137+32=169 

$PTK\,M_3(PO_4)_2=310$ hay $3NTK\, M+190=310$

$\Rightarriow 3NTK\,M=120\\\Rightarrow NTK\,M=40(đvC)$

$\to M$ là $Ca$

{[(37+13):5]-45:5}.7

={[50:5]-9}.7

={10-9}.7

=1.7

=7

a/ $=-x^2+3x+10+x^2-6x+9-6x+12\\=-9x+31$

b/ $=-y^2+y+12+y^2-8y+16-6x+10\\=-13y+38$