Question

cho B=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}\)+\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).số nghiệm của phương trình \(x^3+3Bx^2+27Bx+9B^2=0\) là ?

Answer 1

\(B=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+...+\dfrac{\sqrt{100}-\sqrt{99}}{1}\)

\(=\sqrt{100}-1=9\)

\(x^3+3.9x^2+3.9^2x+9^3=0\)

\(\Leftrightarrow\left(x+9\right)^3=0\)

\(\Leftrightarrow x=-9\)