Bµi cò 3 – 2 = 1 3 – 1 = 2 1 2 – 1 = 3 1 + 2 = Bµi 1. TÝnh: 1 + 1 = 2 - 1 = 2 + 1 = 1 + 2 = 3 - 1 = 3 - 2 = 2 1 3 3 2 1 1 + 2 = 1 + 3 = 1 + 4 = 1 + 1 + 1 = 3 - 1 - 1 = 3 - 1 + 1 = 3 4 5 3 1 3 Bµi 1. TÝnh: 1 + 1 = 2 - 1 = 2 + 1 = 1 + 2 = 2 1 3 3 2 1 1 + 2 = 1 + 3 = 1 + 4 = 1 + 1 + 1 = 3 - 1 - 1 = 3 - 1 + 1 = 3 4 5 3 1 3 3 - 3 3 - 1 1 1 2 = 23 = 32 3 Bµi 1. TÝnh: 1 + 1 = 2 - 1 = 2 + 1 = 1 + 2 = 2 1 3 3 2 1 1 + 2 = 1 + 3 = 1 + 4 = 1 + 1 + 1 = 3 - 1 - 1 = 3 - 1 + 1 = 3 4 5 3 1 3 - 3 3 - 1 2 = 23 = 3 a + b = 3 3 - a = b 3 - b = a 3 -1 2 -2 1 -1 1 +1 3 Trß ch¬i: “§o¸n sè sau hoa” 3 2 2 1 1 2= 2 1 1= 1 2 3= 3 1 2= 2 1 3 = 3 2 1= 1 4 5= 2 2 4= + - + - + - + + Bµi 3: _ + ? 2 1 - = 1 Bµi 4. ViÕt phÐp tÝnh thÝch hîp: a)