Question

[Nguyễn Huy Tú]

Tìm giá trị của x để biểu thức B = \(\frac{x^2-4x+2020}{x^2-2x+1}\)nhận GTNN

@ Quỳnh hentaiz :)) 

Answer 1

[Phan Nghĩa]

\(B=\frac{\left(x-2\right)^2+2016}{\left(x-1\right)^2}=\frac{\left(t-1\right)^2+2016}{t^2}=\frac{t^2-2t+2017}{t^2}\)

\(=1-\frac{2}{t}+\frac{2017}{t^2}=1-2a+2017a^2=2017\left(a^2-2.\frac{1}{4034}+\frac{1}{4034}^2\right)-\frac{2017}{4034^2}+1\)

\(=2017\left(a-\frac{1}{4034}\right)^2+1-\frac{1}{2017^3}\ge1-\frac{1}{2017^3}\)

tự xét dấu = 

Answer 2

[Ĵ❍ƙĔŔᵛᶰ(*•.¸♡ţęąɱ ƒŗęę ƒįŗę❤☆)]

\(B=\frac{\left(x-2\right)^2+2016}{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{\left(t-1\right)^2+2016}{1^2}\)

\(\Leftrightarrow\frac{t^2-2t+2017}{t^2}\)

\(\Leftrightarrow1-\frac{2}{t}+\frac{2017}{t^2}\)

\(\Leftrightarrow1-2a+2017a^2\)

\(\Leftrightarrow a^2-2\times[\frac{1}{4034}+\frac{1^2}{4034}]-\frac{2017}{4034^2}+1\)

\(\Leftrightarrow2017\left(a-\frac{1}{4034}\right)^2+1-\frac{1}{2017}^3\)

phần cuối tự làm nha