Bài tập 9 trang 78 SGK Toán 10 NC

Áp dụng định lý Vi-ét, ta có: 

\(\left\{ {\begin{array}{*{20}{l}}
{{x_1} + {x_2} =  - \frac{b}{a}}\\
{{x_1}.{x_2} = \frac{c}{a}}
\end{array}} \right.\)

Do đó 

\(\begin{array}{*{20}{l}}
{a{x^2} + bx + c = 0 = a\left( {{x^2} + \frac{b}{a}x + \frac{c}{a}} \right)}\\
{ = a\left[ {{x^2} - \left( {{x_1} + {x_2}} \right)x + {x_1}{x_2}} \right]}\\
{ = a\left[ {x\left( {x - {x_1}} \right) - {x_2}\left( {x - {x_1}} \right)} \right]}\\
{ = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)}
\end{array}\)

b) Ta có 

\(\begin{array}{l}
f\left( x \right) =  - 2{x^2} - 7x + 4 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x =  - 4}\\
{x = \frac{1}{2}}
\end{array}} \right.
\end{array}\)

Do đó:

\(\begin{array}{l}
f\left( x \right) =  - 2\left( {x + 4} \right)\left( {x - \frac{1}{2}} \right)\\
 = \left( {x + 4} \right)\left( {1 - 2x} \right)
\end{array}\)

Ta có

\(\begin{array}{l}
g\left( x \right) = \left( {\sqrt 2  + 1} \right){x^2}\\
\,\,\,\,\,\,\,\,\,\, - 2\left( {\sqrt 2  + 1} \right)x + 2 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \sqrt 2 }\\
{x = \frac{{\sqrt 2 }}{{\sqrt 2  + 1}}}
\end{array}} \right.
\end{array}\)

Do đó: 

\(\begin{array}{l}
g\left( x \right) = \left( {\sqrt 2  + 1} \right)\left( {x - \sqrt 2 } \right)\left( {x - \frac{{\sqrt 2 }}{{\sqrt 2  + 1}}} \right)\\
 = \left( {x - \sqrt 2 } \right)\left[ {\left( {\sqrt 2  + 1} \right)x - \sqrt 2 } \right]
\end{array}\)

-- Mod Toán 10 HỌC247