Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
A\left( x \right){\rm{ }} = {\rm{ }}\left( {x + 3} \right) + 2{x^2}\left( {2x{\rm{ }} - {\rm{ }}3} \right) - 2{x^3}\\
= x + 3 + 4{x^3} - 6{x^2} - 2{x^3}\\
= 2{x^3} - 6{x^2} + x + 3\\
B\left( x \right) = - 3x\left( {{x^2} + x - 3} \right){\rm{ }} - {\rm{ }}8 + 8{x^3} - 6x\\
= - 3{x^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 9x - 8 + 8}}{{\rm{x}}^3}{\rm{ - 6x}}\\
{\rm{ = 5}}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 8 }}\\
C\left( x \right) = 4{x^2}{\rm{ }} + {\rm{ }}3{x^3} + 1 - 4x\\
= 3{x^3} + 4{x^2} - 4x + 1\\
a)H\left( x \right){\rm{ }} = {\rm{ }}A\left( x \right){\rm{ }} - {\rm{ }}B\left( x \right){\rm{ }} + C\left( x \right)\;\\
= (2{x^3} - 6{x^2} + x + 3) - ({\rm{5}}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 8 }}) + (3{x^3} + 4{x^2} - 4x + 1)\\
= 2{x^3} - 6{x^2} + x + 3 - 5{x^3} + 3{x^2} - 3x + 8 + 3{x^3} + 4{x^2} - 4x + 1\\
= {x^2} - 6x + 12\\
b)M\left( x \right){\rm{ }} = {\rm{ }}H\left( x \right){\rm{ }} - {\rm{ }}12\\
= {x^2} - 6x + 12 - 12\\
= {x^2} - 6x\\
M(x) = 0 \Leftrightarrow {x^2} - 6x = 0\\
\Leftrightarrow x(x - 6) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 6 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 6
\end{array} \right.
\end{array}\]

Vậy nghiệm pt M(x) là S={0;6}

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Answer 2

\[\begin{array}{l}
A\left( x \right){\rm{ }} = {\rm{ }}\left( {x + 3} \right) + 2{x^2}\left( {2x{\rm{ }} - {\rm{ }}3} \right) - 2{x^3}\\
= x + 3 + 4{x^3} - 6{x^2} - 2{x^3}\\
= 2{x^3} - 6{x^2} + x + 3\\
B\left( x \right) = - 3x\left( {{x^2} + x - 3} \right){\rm{ }} - {\rm{ }}8 + 8{x^3} - 6x\\
= - 3{x^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 9x - 8 + 8}}{{\rm{x}}^3}{\rm{ - 6x}}\\
{\rm{ = 5}}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 8 }}\\
C\left( x \right) = 4{x^2}{\rm{ }} + {\rm{ }}3{x^3} + 1 - 4x\\
= 3{x^3} + 4{x^2} - 4x + 1\\
a)H\left( x \right){\rm{ }} = {\rm{ }}A\left( x \right){\rm{ }} - {\rm{ }}B\left( x \right){\rm{ }} + C\left( x \right)\;\\
= (2{x^3} - 6{x^2} + x + 3) - ({\rm{5}}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 8 }}) + (3{x^3} + 4{x^2} - 4x + 1)\\
= 2{x^3} - 6{x^2} + x + 3 - 5{x^3} + 3{x^2} - 3x + 8 + 3{x^3} + 4{x^2} - 4x + 1\\
= {x^2} - 6x + 12\\
b)M\left( x \right){\rm{ }} = {\rm{ }}H\left( x \right){\rm{ }} - {\rm{ }}12\\
= {x^2} - 6x + 12 - 12\\
= {x^2} - 6x\\
M(x) = 0 \Leftrightarrow {x^2} - 6x = 0\\
\Leftrightarrow x(x - 6) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 6 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 6
\end{array} \right.
\end{array}\]

Vậy nghiệm pt M(x) là S={0;6}

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