qn 9個月以前 a²-2a + b²+4b + c²+2c +d²-2d + 7 ≤ 0 (a²-2a+1) + (b²+4b+4) + (c²+2c+1) + (d²-2d+1) ≤ 0 (a-1)² + (b+2)² + (c+1)² + (d-1)² ≤ 0 由於 (a-1)² + (b+2)² + (c+1)² + (d-1)² ≥ 0 (等號成立條件 a-1 = b+2 = c+1 = d-1 = 0) 故 a=1 , b=-2 , c=-1 , d=1 留言