Tìm các cặp số (x; y) biết: y⁴+ y²+ x²– 8y – 4x + 2xy + 7 = 0.

Hướng dẫn giải

a) (2x – 3)²– 49 = 0

⇔ (2x – 3)²– 7²= 0

⇔ (2x – 3 – 7)(2x – 3 + 7) = 0

⇔ (2x – 10)(2x + 4) = 0

\( \Leftrightarrow \left[ \begin{array}{l}2x - 10 = 0\\2x + 4 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - 2\end{array} \right.\)

Vậy x = 5, x = - 2.

b) 2x(x – 5) – 7(5 – x) = 0

⇔ 2x(x – 5) + 7(x – 5) = 0

⇔ (x – 5)(2x + 7) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x - 5 = 0\\2x + 7 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - \frac{7}{2}\end{array} \right.\)

Vậy \(x = - \frac{7}{2}\), x = 5.

c) x²– 3x – 10 = 0

⇔ x²– 5x + 2x – 10 = 0

⇔ x(x – 5) + 2(x – 5) = 0

⇔ (x – 5)(x + 2) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x - 5 = 0\\x + 2 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - 2\end{array} \right.\)

Vậy x = 5, x = – 2.

Hướng dẫn giải

a) x²– 8x = x(x – 8)

b) x²– xy – 6x + 6y

= x(x – 6) – y(x – 6)

= (x – 6)(x – y)

^{c) x2}– 6x + 9 – y²

= (x²– 6x + 9) – y²

= (x – 3)²– y²

= (x – 3 – y)(x – 3 + y)

d) x³+ y³+ 2x + 2y

= (x³+ y³) + 2(x + y)

= (x + y)(x²– xy + y²) + 2(x + y)

= (x + y)(x²– xy + y²+ 2)