(a^2 - 2ab+b^2)/(a + b)?
828991
(a^2 -2ab + b^2) = (a-b)(a-b) = (a-b)^2\nYou cannot cancel anything, so \n(a^2 -2ab + b^2)/(a+b) does not simplify.\nYou cannot solve for anything because you do not have an equation.
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
402319
ya it is a^2 +b ^2 +2abcos(c)
779257
put x=6ab/(a+b)in the eq(x+3a)/(x-3a) + (x+3b)/(x-3b).\nthen you get(9ab+3a^2/3ab-3a^2)+(9ab+3b^2/3ab-3b^2)\nby bringing the common values out in numerator and denominator and cancelling the them we get\n(3b+a/b-a ) +(3a+b/a-b)\nso the answer is 2
439090
((x^2-2x)/4y)/((x^2-4)/6)\n=(6x(x-2))/(4y*(x-2)(x+2))\n=3x/(2y(x+2))
353701
sqrt(a^2 + b^2) = a + b\nSquare each side.\n[sqrt(a^2 + b^2)]^2 = (a+b)^2\na^2 + b^2 = a^2 + 2ab + b^2\nSubtract a^2 and b^2 from each side\n0 = 2ab  \nDivide both sides by 2\n0 = ab\nEither a = 0 or b = 0 or both equal 0.\nAnswer: a=0 and b > 0, or a > 0 and b=0, or a=0 and b=0\na or b can never be negative(even it one of them is zero).\n\nIn general, the equation \nsqrt(a^2 + b^2) = a + b \nis false.  However, it is true for certain values of a and b.
38332
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
893957
well darling ;\nimagine this like that\n\ny^2 + ( a+b)y + ab \nyou need to find " a+b " & " ab" \nin our function \na+b = 16 & ab = 64 \n( y + a)( y + b) \n( y + 8 ) ( y + 8)\n{ ' 8 + 8 = 16 '   &  ' 8 * 8 = 64 ' }\nso\n( y + 8)^2\n\nGood Luck sweetheart.
924424
The equations:\n(a+b)^2=a^2+2ab+b^2\nA^2-B^2=(A+B)(A-B)\nJames Chan's solutions were right.
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)
706305
R^2=A^2+B^2 -2(A*B)cos(180-acute angle)
714589
x/(x+3) - 2(x-1)/(x+3)(x+2)\n\nx(x+2) - 2 (x-1)\n--------------------\n   (x+3) (x+2)\n\n\nx^2 +2\n----------\n(x+3)(x+2)
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
692175
( x^2 -x +2 )( x^2 + x + 2 )\n\n=>\n a= -1\n b= +2\n c= +1\n d= +2
426775
b = 2g - 10\ng = (b + 10)/2