`1.`\(x^2+2xy+y^2-4=\left(x+y\right)^2-2^2=\left(x+y+2\right)\left(x+y-2\right)\)

`2.`\(3xy^2-12xy+12x=3x\left(y^2-4y+4\right)=3x\left(y-2\right)^2\)

Bổ sung:

Ta có hệ: \(\left\{{}\begin{matrix}23x+62y=1,62\\x+2y=0,06\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,03\\y=0,015\end{matrix}\right.\)

\(\left\{{}\begin{matrix}m_{Na}=0,03.23=0,69\left(g\right)\\m_{Na_2O}=0,015.62=0,93\left(g\right)\end{matrix}\right.\)

Bài 6.

\(S_{Na_2SO_4;10^oC}=\dfrac{7,2}{80}.100=9\left(g\right)\)

Bài 11.

Đặt \(\left\{{}\begin{matrix}n_A=x\\n_{A_2O}=y\end{matrix}\right.\) ( mol )

\(2A+2H_2O\rightarrow2AOH+H_2\)

 x                       x                     ( mol )

\(A_2O+H_2O\rightarrow2AOH\)

 y                          2y      ( mol )

\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)\)

\(2AOH+H_2SO_4\rightarrow A_2SO_4+2H_2O\)

0,06           0,03                                ( mol )

\(\Rightarrow n_{AOH}=x+2y=0,06\) ( mol )

\(\Leftrightarrow x=0,06-2y\)

\(m_{hh}=Ax+y\left(2A+16\right)=1,62\)

\(\Leftrightarrow Ax+2Ay+16y=1,62\)

\(\Leftrightarrow A\left(0,06-2y\right)+2Ay+16y=1,62\)

\(\Leftrightarrow0,06A-2Ay+2Ay+16y=1,62\)

\(\Leftrightarrow y=\dfrac{1,62-0,06A}{16}\)

Vì y là số mol nên \(0< y< 0,03\)

`@`\(y>0\) \(\Rightarrow\dfrac{1,62-0,06A}{16}>0\) \(\Leftrightarrow A< 27\) (1)

`@`\(y< 0,03\) \(\Rightarrow\dfrac{1,62-0,06A}{16}< 0,03\) \(\Leftrightarrow A>19\) (2)

(1);(2) `=>` A là Natri ( Na )

`a.` Ta có: \(\widehat{A}=\widehat{A}_{trong}+\widehat{A}_{ngoài}=8x-9^o+3x-9^o=180^o\) ( kề bù )

\(\Leftrightarrow11x=198^o\)

\(\Leftrightarrow x=18^o\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{D}=2.18^o+9^o=45^o\\\widehat{A}_{trong}=8.18^o-9^o=135^o\end{matrix}\right.\)

\(\Rightarrow\widehat{A}_{trong}+\widehat{D}=135^o+45^o=180^o\)

mà \(\widehat{A}_{trong}\) và \(\widehat{D}\) nằm ở vị trí trong cùng phía

`=>AB////CD`

`=>` Tứ giác ABCD là hình thang

`b.`Ta có: \(\widehat{B}+\widehat{C}=180^o\) ( trong cùng phía )

                Mà  \(\widehat{B}-\widehat{C}=32^o\)

   \(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=106^o\\\widehat{C}=74^o\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\widehat{IBC}=\dfrac{\widehat{B}}{2}=\dfrac{106^o}{2}=53^o\\\widehat{ICB}=\dfrac{\widehat{C}}{2}=\dfrac{74^o}{2}=37^o\end{matrix}\right.\)

Xét tam giác IBC, có:

\(\widehat{IBC}+\widehat{ICB}+\widehat{BIC}=180^o\) ( tổng 3 góc của tam giác )

\(\Rightarrow\widehat{BIC}=180^o-53^o-37^o=90^o\)

`a.`Sơ đồ \(\Leftrightarrow R_1nt\left(R_2\backslash\backslash R_3\right)\)

\(\Rightarrow I=I_1=I_{23}=1,5A\)

\(U_1=I_1.R_1=1,5.3=4,5V\)

\(\Rightarrow U_{23}=U_{AB}-U_1=12-4,5=7,5V\)

\(R_{23}=\dfrac{U_{23}}{I_{23}}=\dfrac{7,5}{1,5}=5\Omega\)

\(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=5\)

\(\Leftrightarrow\dfrac{10.R_3}{10+R_3}=5\)    \(\Leftrightarrow R_3=10\Omega\)

`b.` Sơ đồ \(\Leftrightarrow R_xnt\left(R_2\backslash\backslash R_3\right)\)

\(\Rightarrow I=I_x=I_{23}=0,4A\)

\(U_{23}=I_{23}.R_{23}=0,4.5=2V\)

\(U_x=U_{AB}-U_{23}=12-2=10V\)

\(R_x=\dfrac{U_x}{I_x}=\dfrac{10}{0,4}=25\Omega\)

\(\dfrac{3}{8}+\dfrac{2}{7}+\dfrac{1}{4}=\left(\dfrac{3}{8}+\dfrac{1}{4}\right)+\dfrac{2}{7}=\left(\dfrac{3}{8}+\dfrac{2}{8}\right)+\dfrac{2}{7}=\dfrac{5}{8}+\dfrac{2}{7}=\dfrac{5\times7+2\times8}{56}=\dfrac{51}{56}\)

\(\dfrac{288}{72}=\dfrac{96}{24}=\dfrac{16}{4}=\dfrac{72}{18}\)