b, dk tu xd nhé

\(\Leftrightarrow\dfrac{\left(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\right)\left(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\right)}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-2x=0\)

\(\Leftrightarrow2x\left(\dfrac{1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=1\left(l\right)\end{matrix}\right.\)

ns \(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}>1\)

\(\Rightarrow x=0\left(tm\right)\)

2015 Câu này mik làm rùi tích nha

căn bậc 2 của 13 :3,605551275464...

áp dụng bunhia

\(\left[\left(\sqrt{\dfrac{2}{1-x}}\right)^2+\left(\sqrt{\dfrac{1}{x}}\right)^2\right]\left[\left(\sqrt{1-x}\right)^2+\left(\sqrt{x}\right)^2\right]\)

\(\ge\left(\sqrt{\dfrac{2}{1-x}}.\sqrt{1-x}+\sqrt{\dfrac{1}{x}}.\sqrt{x}\right)^2\)

\(\Leftrightarrow\left(\dfrac{2}{1-x}+\dfrac{1}{x}\right)\left(1\right)\ge\left(\sqrt{2}+\sqrt{1}\right)^2\)

\(\Rightarrow B\ge\left(\sqrt{2}+1\right)^2\)

dấu = xảy ra khi \(\dfrac{\dfrac{2}{1-x}}{1-x}=\dfrac{\dfrac{1}{x}}{x}\Leftrightarrow x=\sqrt{2-1}\)