Toán Lớp 8: tim x biet (3x-2)^3+(2x-3)^3-(5x-5)^3=0
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Toán Lớp 8: tim x biet (3x-2)^3+(2x-3)^3-(5x-5)^3=0
Comments ( 2 )
Phương trình có 3 nghiệm chúc bạn học tốt
ta\,\,co:\\
\left( {3x – 2} \right) + \left( {2x – 3} \right)\\
= 3x – 2 + 2x – 3\\
= 5x – 5\\
\Rightarrow dat\,\left\{ \begin{array}{l}
A = 3x – 2\\
B = 2x – 3
\end{array} \right.\\
\Rightarrow thay\,A + B = 5x – 5\,\,vao\,\,pt\\
\Rightarrow {A^3} + {B^3} – {\left( {A + B} \right)^3} = 0\\
\Rightarrow {A^3} + {B^3} – {A^3} – 3{A^2}B – 3A{B^2} + {B^3} = 0\\
\Rightarrow – 3{A^2}B – 3A{B^2} = 0\\
\Rightarrow {A^2}B + A{B^2} = 0\\
\Rightarrow A.B.\left( {A + B} \right) = 0\\
\Rightarrow \left( {3x – 2} \right).\left( {2x – 1} \right).\left( {5x – 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
3x – 2 = 0\\
2x – 1 = 0\\
5x – 5 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{2}{3}\\
x = \frac{3}{2}\\
x = 1
\end{array} \right.\\
\Rightarrow S \in \{ \frac{2}{3};\frac{3}{2};1\}
\end{array}\]