find the rational expressions x-3 over x+2?
372587
I'm not sure what you are looking for.\n(x-3)/(x+2) does not simplify.\nThe domain is all real numbers except -2.\nThe range is all real numbers except 1.
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
790277
ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
164954
x^3 – 3x^2 + 4x – 12\n= x^2(x – 3) + 4(x – 3)\n= (x-3)(x^2 +4)
786009
-2(x-3) + 4(x+7)\n-2x+6   +4x+28\n2x+34
1019688
(x - 2)*3 = 6 + x\n\n3x - 6 = 6 + x\n\n2x = 12\n\nx = 6
829339
x^2-6x+9 = (x-3)(x-3)\nx^2-9 = (x-3)(x+3)\n\nMultiple everything by (x-3)(x-3)(x+3)\n\n(x+3)(x+3) - (x+2)(x-3) - 5(x-3)(x+3)\n\nx^2+6x+9 - (x^2-x-6) - (5x^2-45)\n\nx^2+6x+9-x^2+x+6-5x^2+45\n\n-5x^2 + 7x + 60
895527
2x^2-8x+3x-12\n2x(x-4)+3(x-4)\n2x+3 = 0 or x-4= 0\nx = -3/2 or x = 4\n\n2x^2+10x-72\n2x^2+18x-8x-72\n2x(x+9)-8(x+9)\nx = 4 or x = -9\n\nx^2+6x-x-6\nx-1 = 0 or x+6 = 0\nx = 1 or x = -6
588927
easy.. -1 , -2 , 1,2,3.\n\n-1 x -2 =2\n2x2x3 = 12
893013
is it x-3/x+2 or (x-3)/(x+2)?\n\nlets assume the second one\n\n2x(x+2)/(x+2)=x-3\nx+2 cancels out\n\n2x=x-3\nx=-3
520679
When dividing rational expressions you flip the second one and multiply\n (2x^2 + 5x)/(9x^2 – 16) ÷ (2x^2-7x + 6)/(3x^2 -x -4)\n= (2x^2 + 5x)/(9x^2 – 16)*(3x^2 -x -4)/(2x^2-7x + 6)\nFactor each one, then you can cancel anything on the top with anything that is on the bottom.
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
98350
limit as x tends to 3 ;( x^2 -3)/2 =(3^2 -3)/2 =(9-3)/2 = 3
672142
it's\n[x(x+2)]/[x*x] -2/3x=\n\n[x+2]/x - 2/3x=\n\n[3[x+2]-2]/3x=\n\n[3x+4]/3x=\n\nx + [4/3]x
872966
1.) 3 &#92;/ x+2 - 4 = 3\nif you mean\n3 sqrt( x+2) -4 =3,\nthen\n3 sqrt(x+2) =7\nsqrt(x+2) = 7/3\nx+2 = 49/9\nx=49/9 -2\n\n2.) &#92;/ x+3 - &#92;/2x-8 = 1\nsqrt(x+3) -sqrt(2x-8) =1\nsqrt(x+3) = sqrt(2x-8) +1\nx+3 = 2x-8 +2 sqrt(2x-8) +1\n-x + 10 = 2 sqrt(2x-8)\nx^2 -20x + 100 = 4 ( 2x-8)\nx^2 - 28x  +132 =0\n'
722836
I think you are asking to simplify the rational expression:\n[(x^3+27)/(9x+27)]/\n[(3x^2-9x+27)/(4x)]\n\nIf this is what you are asking, here's what you do.\n\nWe are dividing by a fraction, so we must flip the second fraction and multiply.  Our problem now becomes:\n[(x^3+27)/(9x+27)]*\n[(4x)/(3x^2-9x+27)]\n\nNow we must factor each expression individually.\n\nThe first expression--x^3+27--is a sum of two cubes and factors as:\n(x+3)(x^2-3x+9)\n\nThe second expression--9x+27--has 9 as a common factor and factors as:\n9(x+3)\n\nThe third expression--4x--cannot be factored, so we leave it alone.\n\nThe fourth expression--3x^2-9x+27--has 3 as a common factor and factors as:\n3(x^2-3x+9)\n\nUsing substitution, our expression:\n[(x^3+27)/(9x+27)]*\n[(4x)/(3x^2-9x+27)]\nbecomes:\n{[(x+3)(x^2-3x+9)]/[9(x+3)]}*\n{(4x)/[3(x^2-3x+9)]}\n\nMultiplying these expressions into one large expression, we get:\n[(x+3)*(x^2-3x+9)*(4x)]/\n[27*(x+3)*(x^2-3x+9)]\n\nCancelling the (x+3)'s and the (x^2-3x+9)'s that are common to the numerator and denominator leaves us with our final answer:\n\n(4x)/27\n\nGood luck.