bài toán khó

C

conga222222

$\begin{array}{l}
neu\;a = 0 \to thoa\;man\\
neu\;a \ne 0 \to \\
\left| b \right| = \left| {a + c} \right| \leftrightarrow \left[ \begin{array}{l}
b = a + c\\
b = - a - a
\end{array} \right.\\
neu\;b = a + c\\
\to a{x^2} + bx + c = 0 \leftrightarrow a{x^2} + \left( {a + c} \right)x + c = 0\\
\leftrightarrow \left( {ax + a} \right)\left( {x + \frac{c}{a}} \right) = 0\\
neu\;b = - a - c\\
\to a{x^2} - \left( {a + c} \right)x + c = 0 \leftrightarrow \left( {ax - a} \right)\left( {x - \frac{c}{a}} \right) = 0\\
\to thoan\;man
\end{array}$
 
K

khangcvp

phuong trinh bac hai

Phuong trinh a x^2+b x+c co nghiem hưu ti <=>b^2-4ac la binh phương mot so huu ti
That vay b^2-4ac=(a+c)^2-4ac=(a-c)^2=>dpcm
 
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