Toán Lớp 8: Tìm các số a,b,c biết
1,-3x^3.(2ax^2-bx+c)=-6x^5+9x^4-3x^3
2,(ax+b)(x^2-cx+2)=x^3+x^2-2
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Toán Lớp 8: Tìm các số a,b,c biết
1,-3x^3.(2ax^2-bx+c)=-6x^5+9x^4-3x^3
2,(ax+b)(x^2-cx+2)=x^3+x^2-2
Giúp mình với ạ ạ ạ !
Comments ( 1 )
1)\\
– 3{x^3}\left( {2a.{x^2} – bx + c} \right)\\
= – 6a.{x^5} + 3b.{x^4} – 3c.{x^3}\\
= – 6{x^5} + 9{x^4} – 3{x^3}\\
\Leftrightarrow \left\{ \begin{array}{l}
– 6a = – 6\\
3b = 9\\
– 3c = – 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = 3\\
c = 1
\end{array} \right.\\
Vậy\,a = 1;b = 3;c = 1\\
2)\\
\left( {a.x + b} \right)\left( {{x^2} – c.x + 2} \right)\\
= a.{x^3} – a.c.{x^2} + 2a.x + b.{x^2} – bc.x + 2b\\
= a.{x^3} + \left( { – ac + b} \right).{x^2} + \left( {2a – bc} \right).x + 2b\\
= {x^3} + {x^2} – 2\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
– a.c + b = 1\\
2a – bc = 0\\
2b = – 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
– c + b = 1\\
2 – bc = 0\\
b = – 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 1;b = – 1\\
c = b – 1 = – 2\\
bc = – 2
\end{array} \right.\\
Vậy\,a = 1;b = – 1;c = – 2
\end{array}$