Đáp án:
Bài `1`
`a, (5x - 1)^2 - (5x - 4)(5x + 4) = 7`
`⇔ (5x)^2 - 2.5x.1 + 1^2 - [(5x)^2 - 4^2] = 7`
`⇔ 25x^2 - 10x + 1 - (25x^2 - 16) - 7 = 0`
`⇔ 25x^2 - 10x + 1 - 25x^2 + 16 - 7 = 0`
`⇔ (25x^2 - 25x^2) - 10x + (1 + 16 - 7) = 0`
`⇔ - 10x + 10 = 0`
`⇔ 10(1 - x) = 0`
`⇔ 1 - x = 0`
`⇒ x = 1`
Vậy `x = 1`
`b, (4x - 1)^2 - (2x + 3)^2 + 5(x + 2)^2 + 3(x - 2)(x + 2) = 500`
`⇔ (4x)^2 - 2.4x.1 + 1^2 - [(2x)^2 + 2.2x.3 + 3^2] + 5(x^2 + 2.x.2 + 2^2) + 3(x^2 - 2^2) = 500`
`⇔ 16x^2 - 8x + 1 - (4x^2 + 12x + 9) + 5(x^2 + 4x + 4) + 3(x^2 - 4) = 500`
`⇔ 16x^2 - 8x + 1 - 4x^2 - 12x - 9 + 5x^2 + 20x + 20 + 3x^2 - 12 - 500 = 0`
`⇔ (16x^2 - 4x^2 + 5x^2 + 3x^2) + (-8x - 12x + 20x) + (1- 9 + 20 - 12 - 500) = 0`
`⇔ 20x^2 - 500 = 0`
`⇔ 20(x^2 - 25) = 0`
`⇔ x^2 - 5^2 = 0`
`⇔ (x - 5)(x + 5) = 0`
`⇒` $\left[\begin{matrix} x-5 = 0\\ x + 5=0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = 5\\ x = -5\end{matrix}\right.$
Vậy `x = 5` hoặc `x = -5`
`c, x^2 + y^2 + 4x - 10y + 29 = 0`
`⇔ x^2 + 4x +4 + y^2 - 10y + 25 = 0`
`⇔ (x^2 + 4x + 4) + (y^2 - 10y + 25) = 0`
`⇔ (x^2 + 2.x.2 + 2^2) + (y^2 - 2.y.5 + 5^2)= 0`
`⇔ (x + 2)^2 + (y - 5)^2 + 0`
`⇒` $\left[\begin{matrix} (x + 2)^2 = 0\\ (y - 5)^2=0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x + 2 = 0\\ y - 5=0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = -2\\ y = 5\end{matrix}\right.$
Vậy `x = -2, y = 5`
`d, - x^2 + 8x - 4y^2 + 20y = 41`
`⇔ -x^2 + 8x - 4y^2 + 20y - 41 = 0`
`⇔ - x^2 + 8x - 16 - 4y^2 + 20y - 25 =0`
`⇔ - (x^2 - 8x + 16) - (4y^2 - 20y + 25) = 0`
`⇔ - (x^2 - 2.x.4 + 4^2) - [(2y)^2 - 2.2y.5 + 5^2] = 0`
`⇔ - (x - 4)^2 - (2y - 5)^2 = 0`
`⇔ - [(x - 4)^2 + (2y - 5)^2 ]= 0`
`⇔ (x - 4)^2 + (2y - 5)^2 = 0`
`⇒` $\left[\begin{matrix} (x - 4)^2 = 0\\ (2y - 5)^2 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x - 4 = 0\\ 2y - 5 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = 4\\ y = \dfrac{5}{2}\end{matrix}\right.$
Vậy `x = 4, y = 5/2`
