ai giúp mình với ạ, mình hứa vote 5 sao ạ

Đáp án:

h) \(\left[ \begin{array}{l}
{x_2} =  - 2\\
{x_2} = 4
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)Thay:x = 3\\
Pt \to {2.3^2} - \left( {m + 3} \right).3 + m - 1 = 0\\
 \to  - 2m + 8 = 0\\
 \to m = 4\\
Thay:m = 4\\
Pt \to 2{x^2} - 7x + 3 = 0\\
 \to \left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{2}
\end{array} \right.\\
b)Thay:x = 2\\
Pt \to 4m - 2\left( {m + 2} \right) + m - 1 = 0\\
 \to 3m - 5 = 0\\
 \to m = \dfrac{5}{3}\\
Thay:m = \dfrac{5}{3}\\
Pt \to \dfrac{5}{3}{x^2} - \dfrac{{11}}{3}x + \dfrac{2}{3} = 0\\
 \to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{1}{5}
\end{array} \right.\\
c)Thay:x = 2\\
Pt \to 4\left( {m + 3} \right) + 4\left( {3m + 1} \right) + m + 3 = 0\\
 \to 17m + 19 = 0\\
 \to m =  - \dfrac{{19}}{{17}}\\
Thay:m =  - \dfrac{{19}}{{17}}\\
Pt \to \dfrac{{32}}{{17}}{x^2} - \dfrac{{80}}{{17}}x + \dfrac{{32}}{{17}} = 0\\
 \to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
d)Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} =  - \dfrac{m}{{4 - m}}\\
{x_1}{x_2} = \dfrac{{1 - m}}{{4 - m}}
\end{array} \right.\\
Thay:{x_1} = 1\\
 \to \left\{ \begin{array}{l}
1 + {x_2} =  - \dfrac{m}{{4 - m}}\\
{x_2} = \dfrac{{1 - m}}{{4 - m}}
\end{array} \right.\left( {DK:m \ne 4} \right)\\
 \to \left\{ \begin{array}{l}
1 + \dfrac{{1 - m}}{{4 - m}} =  - \dfrac{m}{{4 - m}}\\
{x_2} = \dfrac{{1 - m}}{{4 - m}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{4 - m + 1 - m + m}}{{4 - m}} = 0\\
{x_2} = \dfrac{{1 - m}}{{4 - m}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{5 - m}}{{4 - m}} = 0\\
{x_2} = \dfrac{{1 - m}}{{4 - m}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m = 5\\
{x_2} = 4
\end{array} \right.\\
e)Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{4}{{2m - 1}}\\
{x_1}{x_2} = \dfrac{{4m - 3}}{{2m - 1}}
\end{array} \right.\left( {DK:m \ne \dfrac{1}{2}} \right)\\
Thay:{x_1} =  - 1\\
 \to \left\{ \begin{array}{l}
 - 1 + {x_2} = \dfrac{4}{{2m - 1}}\\
 - {x_2} = \dfrac{{4m - 3}}{{2m - 1}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
{x_2} =  - \dfrac{{4m - 3}}{{2m - 1}}\\
 - 1 - \dfrac{{4m - 3}}{{2m - 1}} = \dfrac{4}{{2m - 1}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
{x_2} =  - \dfrac{{4m - 3}}{{2m - 1}}\\
\dfrac{{ - 2m + 1 - 4m + 3 - 4}}{{2m - 1}} = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m = 0\\
{x_2} =  - 3
\end{array} \right.\\
f)Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{ - 1}}{{m - 4}}\\
{x_1}{x_2} = \dfrac{{{m^2} - 4m + 1}}{{m - 4}}
\end{array} \right.\\
Thay:{x_1} =  - 1\\
 \to \left\{ \begin{array}{l}
 - 1 + {x_2} = \dfrac{{ - 1}}{{m - 4}}\\
 - {x_2} = \dfrac{{{m^2} - 4m + 1}}{{m - 4}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
{x_2} =  - \dfrac{{{m^2} - 4m + 1}}{{m - 4}}\\
 - 1 - \dfrac{{{m^2} - 4m + 1}}{{m - 4}} = \dfrac{{ - 1}}{{m - 4}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{ - m + 4 - {m^2} + 4m - 1 + 1}}{{m - 4}} = 0\\
{x_2} =  - \dfrac{{{m^2} - 4m + 1}}{{m - 4}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
 - {m^2} + 3m + 4 = 0\\
{x_2} =  - \dfrac{{{m^2} - 4m + 1}}{{m - 4}}
\end{array} \right.\left( {DK:m \ne 4} \right)\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m = 4\left( l \right)\\
m - 1
\end{array} \right.\\
{x_2} = \dfrac{6}{5}
\end{array} \right.\\
h)Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 2\\
{x_1}{x_2} = {m^2} - 3m
\end{array} \right.\\
Thay:{x_1} = 0\\
 \to \left\{ \begin{array}{l}
{x_2} = 2m - 2\\
{m^2} - 3m = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
m = 0\\
m = 3
\end{array} \right. \to \left[ \begin{array}{l}
{x_2} =  - 2\\
{x_2} = 4
\end{array} \right.
\end{array}\)