.If x = 1+√2 , find the value of (𝑥 −
1
𝑥
)
3
.
23*x^3-23*x^2+6*x-1 = 0
6/2(x+1)
1320-3x=y
show that the lowest poinmt on the curve has coordinates 1,5 from equation y=-2X+4+3
249 divided by 58
57.86×100/867.86
57.86×100
867.86
2x²+y²+4x+4y=0
1 over 15 plus 2 over 5