Đưa các phương, trình sau về dạng \[a{x^2} + bx + c = 0\] chỉ rõ các hệ số \[a,{\rm{ }}b,{\rm{ }}c\]

a)\(5{x^2} + 2x = 4 - x\);                     

b) \({\rm{ }}\frac{3}{5}{x^2} + 2x - 7 = 3x + \frac{1}{2}\)      .

c)\(2{x^2} + x - \sqrt 3 = \sqrt 3 x + 1{\mkern 1mu} \)

d) \(2{x^2} - 2(m - 1)x + {m^2} = 0\), \(m\) là một hằng số

\({\rm{a) }}5{x^2} + 2x = 4 - x \Leftrightarrow 5{x^2} + 3x - 4 = 0\,\)

\[a = 5\];\[b = 3\]; \[c =  - 4\]

b) \(\frac{3}{5}{x^2} + 2x - 7 = 3x + \frac{1}{2}\)

\(\frac{3}{5}{x^2} - x - \frac{{15}}{2} = 0\)

\(a = \frac{3}{5},{\mkern 1mu} {\mkern 1mu} b =  - 1,{\mkern 1mu} {\mkern 1mu} c =  - \frac{{15}}{2}\)

\({\rm{c}}){\rm{ }}2{x^2} + x - \sqrt 3  = \sqrt 3 x + 1\)

\(\begin{array}{l}2{x^2} + (1 - \sqrt 3 )x - \sqrt 3  - 1 = 0\\{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} a = 2;b = 1 - \sqrt 3 ,c =  - \sqrt 3  - 1\end{array}\)

\(\begin{array}{l}{\rm{d}}){\mkern 1mu} {\rm{ }}2{x^2} - 2(m - 1)x + {m^2} = 0\\{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} a = 2,\,\,b =  - 2(m - 1),{\mkern 1mu} \,\,c = {m^2}\end{array}\)