Toán Lớp 10: Giải phương trình
x ²+x-2
———– -(x-2)=0
2x+1
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Toán Lớp 10: Giải phương trình
x ²+x-2
———– -(x-2)=0
2x+1
Comments ( 2 )
⇔ x^2 + x – 2 = 2x^2 -3x -2
{\dfrac{{{x^2} + x – 2}}{{2x + 1}} – \left( {x – 2} \right) = 0}\\
{{\rm{\;}} \Leftrightarrow \dfrac{{{x^2} + x – 2}}{{2x + 1}} = x – 2}\\
{{\rm{\;}} \Leftrightarrow \dfrac{{{x^2} + x – 2}}{{2x + 1}} = \dfrac{{\left( {x – 2} \right)\left( {2x + 1} \right)}}{{2x + 1}}}\\
{ \Leftrightarrow {x^2} + x – 2 = \left( {x – 2} \right)\left( {2x + 1} \right)}\\
{ \Leftrightarrow {x^2} + x – 2 = 2{x^2} + x – 4x – 2}\\
{ \Leftrightarrow {x^2} + x – 2{x^2} – x + 4x = {\rm{ \;}} – 2 + 2}\\
{{\rm{\;}} \Leftrightarrow {\rm{ \;}} – {x^2} + 4x = 0}\\
{ \Leftrightarrow {\rm{ \;}} – x\left( {x – 4} \right) = 0}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{\rm{\;}} – x = 0}\\
{x – 4 = 0}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{x = {\rm{ \;}}4}
\end{array}} \right.}\\
{ \Rightarrow S \in \{ 0;4\} {\rm{\;}}}
\end{array}\]