Question

二次根式加减计算 求大虾结果把步骤都写下来吧！！

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Answer 1

解：原式={[-b+√(b²-4ac)]/(2a)}²+{[-b-√(b²-4ac)]/(2a)]}²
={[(-b)²+2×(-b)×√(b²-4ac)+[√(b²-4ac)]²}/(2a)²+{[(-b)²+2×(-b)×[-√(b²-4ac)]+[-√(b²-4ac)]²}/(2a)²
=[b²-2b√(b²-4ac)+b²-4ac]/(4a²)+[b²+2b√(b²-4ac)+b²-4ac]/(4a²)
=[b²-2b√(b²-4ac)+b²-4ac+b²+2b√(b²-4ac)+b²-4ac]/(4a²)
=(4b²-8ac)/(4a²)
=(b²-2ac)/(a²)

Answer 2

={[根号(b^2-4ac)-b]^2+[根号(b^2-4ac)+b]^2}/(4a^2)
=2*(b^2-4ac+b^2)/(4a^2)
=(ba^2-2ac)/a^2