Question

tìm min,max của \(A=\frac{4x+3}{x^2+1}\) và Max \(B=\frac{4Y^2}{X^2-2XY+3Y^2}\)

Answer 1

\(A=\frac{4x+3}{x^2+1}\)\(=\dfrac{x^2+4x+4-\left(x^2+1\right)}{x^2+1}\)\(=\dfrac{\left(x+2\right)^2}{x^2+1}-\dfrac{x^2+1}{x^2+1}\)\(\dfrac{\left(x+2\right)^2}{x^2+1}-1 \ge -1 \forall x \in \mathbb{R}\)

Dấu "=" xảy ra khi \(x+2=0\Leftrightarrow x=-2\)

Vậy \(A_{min}=-1\Leftrightarrow x=-2\)

\(A=\frac{4x+3}{x^2+1}\)\(=\dfrac{4\left(x^2+1\right)-\left(4x^2-4x+1\right)}{x^2+1}\)\(=4-\dfrac{(2x-1)^2}{x^2+1} \le 4 \forall x \in \mathbb{R}\)

Dấu "=" xảy ra khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)

Vậy \(A_{max}=4\Leftrightarrow x=\frac{1}{2}\)

Answer 2

ĐKXĐ: \(x^2+y^2\ne0\)

\(B=\frac{4y^2}{x^2-2xy+y^2+2y^2}=\frac{4y^2}{\left(x-y\right)^2+2y^2}\) Đạt giá trị lớn nhất khi \(\left(x-y\right)^2+2y^2\) bé nhất \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y\ne0\end{matrix}\right.\)\(\Leftrightarrow x=y\ne0\)

\(\Rightarrow B_{Max}=\frac{4y^2}{2y^2}=2\)

Vậy \(B_{max}=2\Leftrightarrow x=y\ne0\)