Question

求解几道分式证明题大神们帮帮忙

1，已知a^2+c^2=2 b^2,求证1/a+b + 1/b+c = 2/a+c.提示：分析法，将结论去分母展开 2，设y + z/ay+bz = z + x/az+bx = x+y/ax+by,求证a=b或x=y=z。提示：设参法，y+z/ay+bz=1/k

Answer 1

1：1/a+b + 1/b+c ；2/a+c.两式同乘以（a+b)(a+c)(b+c) 得1/a+b + 1/b+c =（a+c)(b+c)+（a+b)(a+c)=a^2+c^2+2ab+2ac+2bc 2/a+c.=2（a+b)(b+c)=2ab+2ac+2bc+2 b^2 因为a^2+c^2=2 b^2 所以1/a+b + 1/b+c = 2/a+c 2：y + z/ay+bz = z + x/az+bx = x+y/ax+by同乘以axyz 得axy^2z+xz^2+abz^2xy=axyz^2+x^2y+abx^2yz=ax^2yz+y^2z+abxy^2z 因高册卖为等式恒成立，所以abz^2xy=abx^2yz=abxy^2z；axy^2z=axyz^2=ax^2yz；戚逗姿悔xz^2=x^2y=y^2z 所以x=y=z 另外我不知道设参法迮么用.....

Answer 2

1、 a^2+c^2 = 2b^2 a^2+c^2+2ab+2bc+2ca = 2b^2+2ab+2bc+2ca (a+c)^2+2b(a+c) = 2(a+b)(b+c) (a+c+2b)(a+c) = 2(a+b)(b+c) (a+c+2b)/((a+b)(b+c)) = 2/(a+c) (b+c)/((a+b)(b+c)) + (a+b)/((a+b)(b+c)) = 2/(a+c) 1/(a+b) + 1/(b+c) = 2/(a+c) 2、清激 y + z/ay+bz = z + x/轮友az+bx = x+y/ax+by y + z/ay+bz = z + x/az+bx (y+z)(az+bx) = (z+x)(ay+bz) ayz+bxy+az^2+bxz = ayz+axy+bz^2+bxz bxy+az^2 = axy+bz^2 (a-b)(z^2-xy) = 0 同理(a-b)(x^2-yz) = 0 ，答桐袜(a-b)(y^2-xz) = 0 (a-b)(z^2-xy) + (a-b)(x^2-yz) + (a-b)(y^2-xz) = 0 (a-b)(z^2-xy+x^2-yz+y^2-xz) = 0 (a-b)(2z^2-2xy+2x^2-2yz+2y^2-2xz) = 0 (a-b)((x-y)^2+(y-z)^2+(z-x)^2) = 0 a-b = 0 或 (x-y)^2+(y-z)^2+(z-x)^2 = 0 a=b 或 x=y=z