Toán Lớp 8: Tìm các cặp số (x; y) biết: y 4 + y 2 + x 2 – 8y – 4x + 2xy + 7 = 0.

y⁴+ y²+ x²– 8y – 4x + 2xy + 7 = 0

⇔ y⁴– 2y²+ 1 + 2y²– 4y + 2 + x²+ 2xy + y²– 4x – 4y + 4 = 0

⇔ (y²– 1)²+ 2(y – 1)²+ (x + y)²– 4(x + y) + 4 = 0

⇔ (y²– 1)²+ 2(y – 1)²+ (x + y – 2)²= 0

Vì \({\left( {{y^2} – 1} \right)^2} \ge 0\forall y\)

\(2{(y – 1)^2} \ge 0\forall y\)

\({(x + y – 2)^2} \ge 0\forall x,y\)

⇒ (y²– 1)²+ 2(y – 1)²+ (x + y – 2)²≥ 0

Dấu bằng xảy ra khi \(\left\{ {\begin{array}{*{20}{l}}{{{\left( {{y^2} – 1} \right)}^2} = 0}\\{2{{(y – 1)}^2} = 0}\\{{{(x + y – 2)}^2} = 0}\end{array}} \right.\)\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{y^2} – 1 = 0}\\{y – 1 = 0}\\{x + y – 2 = 0}\end{array}} \right.\)\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{y^2} – 1 = 0}\\{y – 1 = 0}\\{x + y – 2 = 0}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = \pm 1}\\{y = 1}\\{x + y = 2}\end{array}} \right.\)

\( \Rightarrow x = y = 1\)

Vậy cặp số cần tìm (x; y) là (1; 1).