Open mjj42 opened 1 month ago
https://blog.mywshfdz.top/2024/09/09/%E9%9F%A6%E8%BE%BE%E5%AE%9A%E7%90%86/?
韦达定理证明 [ax^2+bx+c=0\] [a(x^2+\frac{b}{a}x)+c=0\] [a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c=0\] [a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\] [(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\]
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1 month ago mjj42
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1 month ago
https://blog.mywshfdz.top/2024/09/09/%E9%9F%A6%E8%BE%BE%E5%AE%9A%E7%90%86/?
韦达定理证明 [ax^2+bx+c=0\] [a(x^2+\frac{b}{a}x)+c=0\] [a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c=0\] [a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\] [(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\]