how do I solve 1+x/1-x > 2-x/2+x ?? please explain me how to solve more of these kinds.?
925554
take all terms over to one side:\n(1+x)/(1-x) - (2-x)/(2+x) > 0\n\nfind a common denominator:\n[ (1+x)(2+x) - (2-x)(1-x) ] / [ (1-x)(2+x) ] > 0\n\nby rearranging the top line, we get:\n6x / [ (1-x)(2+x) ] > 0\n\nfrom this we can see the critical values are -2, 0, 1.\ndraw up a table (5 by 5):\n[space]             (x < -2)      (-2 < x < 0)     (0 < x < 1)     (1 < x)\n6x            -ve            -ve              +ve            +ve\n1-x          +ve           +ve              +ve            -ve\n2+x          -ve           +ve              +ve            +ve\noverall     +ve           -ve               +ve            -ve\n\n[table headings are determined by critical values.\nto find whether 6x is +ve or -ve for x < -2, substitute in a possible value of x eg. -3 --> 6(-3) = -18 therefore -ve.\nto find whether it is overall +ve or -ve, add the signs in that column, eg. for x < -2, a -ve plus a +ve = -ve, add on another -ve and it's +ve.]\n\nsince the identity asks for > 0, we want the positive results ie.\nx < -2 and 0 < x < 1.\n\nhope that helps :)\nand hopefully the table makes sense...
103674
2x^2 - x = 0  { by adding 1 to both sides }\nx^2 - (1/2)x = 0  { by dividing both sides by 2 }\nx (x - 1/2) = 0 { by factoring }\n\nsolves when either x = 0 or x = 1/2 . QED
350975
-2 * x -2 * 6 > x +4\n-2x - 12 > x + 4\n-2x - 12 + 12 > x + 4 + 12\n-x - 2x > x - x + 16\n-3x > 16\n-3/-3x > 16/-3 flip the greater than less than for the negative\n\nx< -16/3\nx<-5.3333
664470
Since 1/(a+b+x)=1/a+1/b+1/x\nSo 1/(a+b+x) - 1/x =1/a+1/b\nSo (x-a-b-x)/(x(a+b+x)) = (a+b)/ab\nSo -(a+b)/(a+x+b)=(a+b)/ab\nSo -1/(x(a+b+x))= 1/ab\nSo x(a+b+x) = - ab\nSo x^2+ (a+b)*x + ab =0\nAnalyzing :\n(x+a)(x+b)=0\nEither x = -a\nOr     x + -b
159688
Is the problem as it is written or is it \n1/(x+2)+2/(x-2)=1/([x^2]-4) ?????\nThe parenthises denote that the denominator includes everything in parenthises.\n\n\nI'm guessing that it is the second form.  In this case we find a least common denominator.  \nSince(x^2-4)=(x+2)(x-2)    we must do the following multiplication  [1/(x+2)*(x-2)/(x-2)] +[2/(x-2)*(x+2)/(x+2)]=1/(x^2+4)\nBy simplifying multiplication and getting all terms This gives us (x-2+2x+4-1)/(x^2-4)=0  \nSince the denominator CAN NOT =0 we have to be sure that x IS NOT =+ or - 2 since this is when x^2-4=0\nSo simplify the numerator by combining like terms we get\n(3x+1)/(x^2-4)=0  \nSince 0 divided by anything except 0 is always 0 we set the numerator equal to 0 and solve for x.\n3x+1=0    \n3x=-1    \nx=-1/3
829201
x+1 / 2x = 2/3\nmultiply both sides by 2x; giving:\nx+1 = 4x/3\nmultiply both sides by 3; giving:\n3x +3 = 4x\nsubtract 3x from both sides; giving:\n3 = x\n____________\n9/2x-1 = 3/x\nmultiply both sides by 2x-1; giving:\n9 = (6x-3)/x\nmultiply both sides by x; giving\n9x = 6x-3\nsubtract 6x from both sides; giving:\n3x = -3\ndivide both sides by 3; giving:\nx = -1
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
732332
1/(1*2) = 1/2\n1/(1*2) + 1/(2*3) = 1/2 + 1/6 = 2/3\n1/(1*2) + .... + 1/(3*4) = 2/3 + 1/12 = 3/4\n\nA pattern is starting to appear eh??\n\nLets see:\n\n1/(1*2) + .... + 1/(4*5) = 3/4 + 1/20 = 4/5 Yesssssssssssss!!\n\nSo I suggest that 1/(1*2) + 1/(2*3) + ..... + 1/(99*100) = 99/100\n\n(See also Scott's formal proof above\n\nHe has based it on the fact that 1/[(n - 1)n] = 1/(n - 1) - 1/n\n\nie 1/4*5 = 1/4 - 1/5)
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
602540
x + 3 >  8 ==> x >  8 - 3  ==> x > 5\nx - 4 < -2 ==> x < -2 + 4 ==> x < 2\n\nAnswer... not possible.
554172
Um...\n\n2(-X) + 3(-X) + X + 8\n\n-2X - 3X + X + 8\n\n-5X + X + 8\n\n-4X + 8\n\n8 - 4X
733649
(4)/(x+7) + (x+3)/(x+7)\n=(4 + (x+3))/(x+7)\n=(4 + x + 3)/(x+7)\n=( x +7)/(x+7)\n=1
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
770190
x² / √(1-x) \n= x² √(1-x)/(1 - x)\n= (-1 + 1 - x²)√(1-x)/(1 - x)\n= (1 - x²)√(1-x)/(1 - x) - √(1-x)/(1 - x)\n= (1 + x)√(1-x) - √(1-x)/(1 - x)\n= x√(1-x) + √(1-x) - 1/√(1-x)\n\nSo ∫x² / √(1-x) . dx\n= ∫[x√(1-x) + √(1-x) - 1/√(1-x)] . dx\n\nNow ∫x√(1-x)dx \n= x * -1 * 2/3 (√(1-x)³ - ∫[-1 * 2/3 (√(1-x)³ . 1)dx\n= -2x/3 * √(1-x)³ + 2/3∫(√(1-x)³ dx\n= -2x/3 * √(1-x)³ + 2/3*2/5 * -1 * √((1-x)^5)\n= -2x/3 * √(1-x)³ - 4/15 * √((1-x)^5)\n\n∫√(1-x) dx = \n\n∫1/√(1-x) dx = -2 *√(1-x)\n\nSo ∫x² / √(1-x) . dx\n= ∫[x√(1-x) + √(1-x) - 1/√(1-x)] . dx\n=  -2x/3 * √(1-x)³ - 4/15 * √((1-x)^5) - 2/3 * √(1-x)³ + 2 *√(1-x) + c\n= - 4/15 * √((1-x)^5) -2/3 (x + 1)√(1-x)³ + 2 *√(1-x)  + c
909284
((1/x) + (1/y)) / ((1/x) - (1/y))\n\n((y + x)/xy) / ((y - x)/xy)\n\n((y + x)/xy) * ((xy)/(y - x))\n\n(xy(y + x))/(xy(y - x))\n\n(y + x)/(y - x) = (x + y)/(-x + y) = (x + y)/(-(x - y)) = -(x + y)/(x - y)
136700
8/x - 3/(x-1) =1\n1. Multiple x on both sides.\n8 - 3x/(x-1) = x\n\n2. Multiple (x-1) on both sides\n8(x-1) - 3x = x (x-1)\n\n3. Simplify\n8x - 8 - 3x = x^2 - x\nx^2 - 6x + 8 = 0\n\n4. Solve\n(x - 4)(x - 2) = 0\nx = 4 or x = 2
751073
Oh why you asked same question 2 times ???? \nAny way, the answer is what i explained for you earlier\n\nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\n\nGood Luck