Question

Cho ABC=2017.Tính giá trị của K=\(\frac{a}{ab+a+2017}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)

Tìm giá trị lớn nhất của H=\(\frac{3-4x}{x^2+1}\)

Answer 1

\(K=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017bc}{abc+2017bc+2017b}\)

\(K=\frac{a}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{2017bc}{2017\left(bc+b+1\right)}\)

\(K=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=\frac{bc+b+1}{bc+b+1}=1\)

\(H=\frac{4x^2+4-4x^2-4x-1}{x^2+1}=\frac{4\left(x^2+1\right)}{x^2+1}-\frac{\left(2x+1\right)^2}{x^2+1}=4-\frac{\left(2x+1\right)^2}{x^2+1}\le4\)

\(\Rightarrow H_{max}=4\) khi \(x=-\frac{1}{2}\)