a)

\(\left\{{}\begin{matrix}4x+5y=3\\x-3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+5y=3\\4x-12y=20\end{matrix}\right.\\ \Leftrightarrow17y=-17\Leftrightarrow y=-1\Leftrightarrow x=5+3.\left(-1\right)=2\)

b)

\(\left\{{}\begin{matrix}7x-2y=1\\3x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-2y=1\\6x+2y=12\end{matrix}\right.\\ \Leftrightarrow13x=13\Leftrightarrow x=1\Leftrightarrow y=\dfrac{7.1-1}{2}=3\)

Còn c, d cứ làm tương tự nha

\(n_{MnO_2}=\dfrac{69,6}{87}=0,8\left(mol\right)\)

\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+2Cl_2+2H_2O\)

0,8------------------------------>1,6

Chất oxi hóa là \(MnO_2\)

\(Mn^{+4}+2e\rightarrow Mn^{+2}\)

Chất khử là HCl

\(Cl^{-1}\rightarrow Cl^0+2e\)

\(n_{KOH}=0,5.4=2\left(mol\right)\)

\(3Cl_2+2KOH\rightarrow KCl+KClO+H_2O\)

0,8--->\(\dfrac{8}{15}\)------->\(\dfrac{4}{15}\)---->\(\dfrac{4}{15}\)

\(CM_{KOH}=\dfrac{2-\dfrac{8}{15}}{0,5}=2,93M\)

\(CM_{KCl}=\dfrac{\dfrac{4}{15}}{0,5}=0,5M\\ CM_{KClO}=\dfrac{\dfrac{4}{15}}{0,5}=0,5M\)

\(C=\dfrac{x^3-6x^2+11x-6}{x^2-3x+2}=\dfrac{x^3-3x^2+2x-3x^2+9x-6}{x^2-3x+2}\\ =\dfrac{x\left(x^2-3x+2\right)-3\left(x^2-3x+2\right)}{x^2-3x+2}=\dfrac{\left(x-3\right)\left(x^2-3x+2\right)}{x^2-3x+2}=x-3\)

\(A=\dfrac{3x^2-11x-14}{4x^2+7x+3}=\dfrac{3x^2+3x-14x-14}{4x^2+4x+3x+3}=\dfrac{3x\left(x+1\right)-14\left(x+1\right)}{4\left(x+1\right)+3\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(3x-14\right)}{\left(x+1\right).\left(4+3\right)}=\dfrac{3x-14}{7}\)

\(B=\dfrac{x^2+13x-7x-91}{x^2+15x-7x-105}=\dfrac{x\left(x+13\right)-7\left(x+13\right)}{x\left(x+15\right)-7\left(x+15\right)}=\dfrac{\left(x-7\right)\left(x+13\right)}{\left(x-7\right)\left(x+15\right)}\\ =\dfrac{x+13}{x+15}\)

ghi phương trình hóa học ra á theo hệ số theo số mol ý 

Gọi x, y là số mol của \(C_2H_5OH\) và \(C_6H_5OH\)

r ghi pthh theo hệ số mol á