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b. Ta có \(\Delta ADE\infty\Delta ABC\left(cgc\right)\)

\(\Rightarrow\dfrac{S_{ADE}}{S_{ABC}}=\left(\dfrac{AD}{AB}\right)^2=\dfrac{AD^2}{AB^2}\)(1)

Xét \(\Delta ADB\) vuông cân tại D có:

\(AD^2+AD^2=AB^2\)(Pitago)

\(\Leftrightarrow2AD^2=AB^2\)\(\Rightarrow\dfrac{AD^2}{AB^2}=\dfrac{AD^2}{2AD^2}=\dfrac{1}{2}\)(2)

Từ (1),(2)\(\Rightarrow\dfrac{S_{ADE}}{S_{ABC}}=\dfrac{1}{2}\)

1+2-3-4+5+6-7-8+9+...+994-995-996+997+998

=(2-3-4+5)+...+(994-995-996+997)+998+1

=0+...+0+999=999

ta có: \(\dfrac{a^2+b^2}{ab}=\dfrac{13}{6}\Rightarrow a^2+b^2=\dfrac{13}{6}ab\)

ta có :\(\left(\dfrac{a+b}{a-b}\right)^2=\left(\dfrac{a^2+b^2+2ab}{a^2+b^2-2ab}\right)=\left(\dfrac{\dfrac{13}{6}ab+2ab}{\dfrac{13}{6}ab-2ab}\right)=25\)

vì \(0< a,b\Rightarrow\dfrac{a+b}{a-b}=5\)

\(A=28x^2-20x+1\)

\(A=\left(\sqrt{28}x-\dfrac{5\sqrt{7}}{7}\right)^2-\dfrac{18}{7}\)

\(A\ge\dfrac{-18}{7}\)(dấu "=" xảy ra\(\Leftrightarrow x=\dfrac{5}{14}\))