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\(x+y\le xy\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\le1\)

\(M=\dfrac{1}{2\left(x^2+y^2\right)+y^2}+\dfrac{1}{2\left(x^2+y^2\right)+x^2}\le\dfrac{1}{4xy+y^2}+\dfrac{1}{4xy+x^2}\)

\(B\le\dfrac{1}{25}\left(\dfrac{4}{xy}+\dfrac{1}{y^2}\right)+\dfrac{1}{25}\left(\dfrac{4}{xy}+\dfrac{1}{x^2}\right)=\dfrac{1}{25}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}+\dfrac{6}{xy}\right)\)

\(M\le\dfrac{1}{25}\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2+\dfrac{3}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\right]=\dfrac{1}{10}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le\dfrac{1}{10}\)

\(M_{max}=\dfrac{1}{10}\) khi \(x=y=2\)

Sử dụng BĐT cộng mẫu:

\(\dfrac{1}{xy}+\dfrac{1}{xy}+\dfrac{1}{xy}+\dfrac{1}{xy}+\dfrac{1}{y^2}\ge\dfrac{\left(1+1+1+1+1\right)^2}{xy+xy+xy+xy+y^2}=\dfrac{25}{4xy+y^2}\)

\(\Rightarrow\dfrac{1}{4xy+y^2}\le\dfrac{1}{25}\left(\dfrac{4}{xy}+\dfrac{1}{y^2}\right)\)

Đáp án C.

Ta có:

G T ⇔ 5 x + 2 y + x + 2 y − 3 − x − 2 y = 5 x y − 1 − 3 1 − x y + x y − 1.

Xét hàm số

f t = 5 t + t − 3 − t ⇒ f t = 5 t ln 5 + 1 + 3 − t ln 3 > 0   ∀ t ∈ ℝ

Do đó hàm số đồng biến trên ℝ  suy ra f x + 2 y = f x y − 1 ⇔ x + 2 y = x y − 1

⇔ x = 2 y + 1 y − 1 ⇒ T = 2 y + 1 y − 1 + y . Do x > 0 ⇒ y > 1  

Ta có:  T = 2 + y + 3 y − 1 = 3 + y − 1 + 3 y − 1 ≥ 3 + 2 3 .

Đáp án C.

Ta có: GT

<=> 5^x+2y + x + 2y – 3^–x–2y = 5^xy–1 – 3^1–xy + xy – 1.

X é t   h à m   s ố   f t = 5 t + t - 3 - t

⇒ f t = 5 t ln 5 + 1 + 3 - t ln 3 > 0   ∀ t ∈ ℝ

Do đó hàm số đồng biến trên  ℝ suy ra

f(x+2y) = f(xy – 1) <=> x+ 2y = xy – 1

⇔ x = 2 y + 1 y - 1 ⇒ T = 2 y + 1 y - 1 + y .

Do x > 0 => y > 1.

Ta có:

T = 2 + y + 3 y - 1 = 3 + y - 1 + 3 y - 1 ≥ 3 + 2 3 .