x³ + y³ + 3(x² + y²) + 4(x + y) + 4 = 0

\( \Leftrightarrow \) (x + y)³ – 3xy(x + y) + 3(x + y)² – 6xy + 4(x + y) + 4 = 0

\( \Leftrightarrow \)[(x + y)³ + 2(x + y)² +(x + y)² + 4(x + y) + 4] – [3xy(x +y) + 6xy] = 0

\( \Leftrightarrow \)[(x + y)²(x + y + 2) + (x + y + 2)²] – 3xy(x + y + 2) = 0

\( \Leftrightarrow \)(x + y + 2)[(x + y)² + x + y + 2] – 3xy(x + y + 2) = 0

\( \Leftrightarrow \) (x + y + 2)[(x + y)² + x + y + 2 – 3xy] = 0

\( \Leftrightarrow \) (x + y + 2)(x² + 2xy + y² + x + y + 2 – 3xy) = 0

\( \Leftrightarrow \) (x + y + 2)[(x² – 2xy + y²) + (x² + 2x + 1) + (y² + 2y + 1) + 2] : 2 = 0

\( \Leftrightarrow \) (x + y + 2)[(x – y)² + (x + 1)² + (y + 1)² + 2] = 0

\( \Leftrightarrow \) x + y + 2 = 0

\( \Leftrightarrow \) x + y = −2

Ta có: \(M = \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{{ - 2}}{{xy}}\)

Vì 4xy ≤ (x + y)²

\( \Leftrightarrow \) 4xy ≤ (−2)²

\( \Leftrightarrow \) 4xy ≤ 4

\( \Leftrightarrow \) xy ≤ 1

\( \Rightarrow \frac{1}{{xy}} \ge 1 \Leftrightarrow \frac{{ - 2}}{{xy}} \le - 2\) (do xy >0)

\( \Leftrightarrow \)M ≤ −2.

Dấu “=” xảy ra khi x = y = −1

Vậy M_max = −2 khi x = y = −1.

Vậy M_min = −2.

So sánh: 50^20 và 2010^10