\(7x^2-13xy-2y^2=0\)

\(\Leftrightarrow7x^2-14xy+xy-2y^2=0\)

\(\Leftrightarrow7x\left(x-2y\right)+y\left(x-2y\right)=0\)

\(\Leftrightarrow\left(7x+y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow x=2y\) (do x;y>0)

Do đó: \(A=\frac{2.2y-6y}{7.2y+4y}=\frac{-2y}{18y}=-\frac{1}{9}\)

\(\sqrt{x+2009}-y^2=\sqrt{y+2009}-x^2\)

<=> \(\left(\sqrt{x+2009}-\sqrt{y+2009}\right)+\left(x^2-y^2\right)=0\)

<=> \(\left(x-y\right)\left(\frac{1}{\sqrt{x+2009}+\sqrt{y+2009}}+x+y\right)=0\)

<=> x - y = 0 vì x; y dương 

<=> x = y 

khi đó: \(A=x^2+2x^2-2x^2+2x+2009=x^2+2x+2009\)

Bạn xem lại đề nhé!

Đáp án C.

Từ giả thiết ta có

  ln x + y + 1 + 3 x + y + 1 = ln 3 x y + 3.3 x y   (*)

Xét  f t = ln t + 3 t  hàm trên  0 ; + ∞ , ta có  f ' t = 1 t + 3 > , ∀ t > 0

Do đó  * ⇔ x + y + 1 = 3 x y ⇔ 3 x y − 1 = x + y ≥ 2 x y ⇔ 3 xy − 2 x y − 1 ≥ 0

Suy ra  x y ≥ 1 ⇒ x y ≥ 1.

Đáp án D

Toán lớp 0 ?????  \(\text{ 🤔 }\text{ 🤔 }\text{ 🤔 }\text{ 😅 }\text{ 😅 }\text{ 😅 }\)

\(3x^2+2y^2=5xy\)

\(\Leftrightarrow3x^2+2y^2-5xy=0\)

\(\Leftrightarrow2\left(x^2-2xy+y^2\right)+x^2-xy=0\)

\(\Leftrightarrow2\left(x-y\right)^2+x\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[2\left(x-y\right)+x\right]=0\)

\(\Leftrightarrow\left(x-y\right)\left(3x-2y\right)=0\)

\(\Leftrightarrow3x-2y=0\Leftrightarrow x=\dfrac{2y}{3}\) Thay vào S

\(\Rightarrow S=\dfrac{y+\dfrac{4y}{3}}{y-\dfrac{4y}{3}}=-7\)