解方程:(8x+24)/(x^2+6x+9)+x^2/(x^2-9)=(x^2-9)/(x^2-6x+9) 1/(x-7
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解方程:(8x+24)/(x^2+6x+9)+x^2/(x^2-9)=(x^2-9)/(x^2-6x+9) 1/(x-7)+1/(x-4)=1/(x-5)+1/(x-6)
a为何值时,方程x/(x-3)=2+a/(x-3)会产生增根?
(1) (8x+24)/(x^2+6x+9)+x^2/(x^2-9)=(x^2-9)/(x^2-6x+9)
(2) 1/(x-7)+1/(x-4)=1/(x-5)+1/(x-6)
(3) a为何值时,方程x/(x-3)=2+a/(x-3)会产生增根?
a为何值时,方程x/(x-3)=2+a/(x-3)会产生增根?
(1) (8x+24)/(x^2+6x+9)+x^2/(x^2-9)=(x^2-9)/(x^2-6x+9)
(2) 1/(x-7)+1/(x-4)=1/(x-5)+1/(x-6)
(3) a为何值时,方程x/(x-3)=2+a/(x-3)会产生增根?
(1)原方程化为 8(x+3)/[(x+3)²]+x²/[(x+3)(x-3)]=[(x+3)(x-3)]/[(x-3)²]
则 8/(x+3)+x²/[(x+3)(x-3)]=(x+3)/(x-3)
两边同时乘 (x+3)(x-3),得 8(x-3)+x²=(x+3)²,即 8x-24+x²=x²+6x+9
解得 x=33/2
(2)两边分别通分,得 (2x-11)/[(x-7)(x-4)]=(2x-11)/[(x-6)(x-5)]
则 (2x-11){1/[(x-7)(x-4)]-1/[(x-6)(x-5)]}=0
所以 2x-11=0 或 1/[(x-7)(x-4)]-1/[(x-6)(x-5)]=0
解得 x=11/2 或 无解
(3)方程的增根只可能是 x=3,
原方程两边乘 (x-3) 得 x=2(x-3)+a
代入 x=3 ,得 3=a
故 a=3
则 8/(x+3)+x²/[(x+3)(x-3)]=(x+3)/(x-3)
两边同时乘 (x+3)(x-3),得 8(x-3)+x²=(x+3)²,即 8x-24+x²=x²+6x+9
解得 x=33/2
(2)两边分别通分,得 (2x-11)/[(x-7)(x-4)]=(2x-11)/[(x-6)(x-5)]
则 (2x-11){1/[(x-7)(x-4)]-1/[(x-6)(x-5)]}=0
所以 2x-11=0 或 1/[(x-7)(x-4)]-1/[(x-6)(x-5)]=0
解得 x=11/2 或 无解
(3)方程的增根只可能是 x=3,
原方程两边乘 (x-3) 得 x=2(x-3)+a
代入 x=3 ,得 3=a
故 a=3
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