3! - [(3 + 3)/3] = 6 - 2 = 4
Or
(√3 x √3) + (3/3) = 3 + 1 = 4
3x7+4x2
(3*3+3)/3=4
4=4(3s) 4=12s s= 1/3
It could be done by: (3*3)/(3*3) = 1
33/3 = 11
(3 + 3) / .3
3 * 3 / ( 3 * 3 ) = 1 but that uses only four 3s, so 33 / ( 3 * 3 * 3 ) = 1 uses five 3s
4 3s = 4*3 = 12, which is a rational number.
You can't unless something is missing in your question.
4y - 4 + 3s + 7 - 6a - 4 = 0 4y + 3s + 7 - 6a - 4 = 4 4y + 3s - 6a - 4 = 4 -7 4y + 3s -6a = 4 - 7 + 4 4y + 3s - 6a = -3 + 4 4y + 3s - 6a = 1 y + 3s - 6a = 1/4 y = -3s + 6a + 1/4 3s = -y + 6a + 1/4 s = -1/3y + 6a + 1/4 -6a = -1/3y + a + 1/4 -a = -1/3y x 1/6 + a + 1/4 x 1/6 a = 1/18y - 1 - 1/24 a = 1/18y - 25/24
4
Use two of the 3s to make 33. Then: 33 - 3 = 11 11 + 3 = 14