Question

Tìm GTNN của A =\(\dfrac{y^2}{9x^2-12xy+5y^2}\)

Answer 1

\(A=\dfrac{y^2}{9x^2-12xy+5y^2}\left(x\ne0\right)\)

\(=\dfrac{y^2}{\left(3x\right)^2-12xy+4y^2+y^2}=\dfrac{y^2}{\left(3x-2y\right)^2+y^2}\)

Ta có : y² ≥ 0 ∀ y

\(\left(3x-2y\right)^2\ge0\)

\(\Rightarrow A\ge0\forall y\) , \(x\ne0\)

=> GTNN của A = 0

Answer 2

\(\dfrac{Y^2}{9x^2-12xy+5y^2}=\dfrac{1}{\dfrac{9x^2}{y^2}-\dfrac{12x}{y}+5}taco:\dfrac{9x^2}{y^2}-\dfrac{12x}{y}+5\)

\(=\left(\dfrac{3x}{y}-2\right)^2+1\le1\)

\(\Rightarrow\dfrac{1}{\dfrac{9x^2}{y^2}-\dfrac{12x}{y}+5}\ge1\)

\(\Rightarrow A\min\limits=1khi\dfrac{3x}{y}=2\)

Answer 3

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Answer 4

yes