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关于x的分式方程1/x%2+k/x+2=4/x^2%4无解,求k的值...

解: 1/(x-2) +k/(x+2)= 4/[(x+2)(x-2)] 去分母,等式两边同乘以(x+2)(x-2) (x+2)+k(x-2)=4 (k+1)x-2(k+1)=0 (k+1)(x-2)=0 k=-1时,等式恒成立,x为任意实数,不满足题意,因此k≠-1 x-2=0 x=2,代入分式方程检验,分母x-2=0,无意义。x=2是分式...

由原方程得[1*(x+2)+k*(x-2)]/[(x-2)(x+2)]=3/[(x-2)(x+2)], x+2+kx-2k=3, (k+1)x=2k+1, x=2-1/(k+1). ∵原方程无解,∴x+2=0或x-2=0,即x=±2.若x=2,则1/(k+1)=0,无解,舍去;若x=-2,则1/(k+1)=4,k=-3/4. 答:k=-3/4.

方程(x-2)/(x+2)+k/(x²-4)=(x+2)/(x-2) 两边同时乘以x²-4即(x+2)(x-2) (x-2)²+k=(x+2)² 化简得8x=k,x=k/8 若原方程无解,那么x=k/8是方程的增根 则k/8=2或k/8=-2 ∴k=16或k=-16

2/(x-2)+mx/(x²-4)=3/(x+2) 去分母,得: 2(x+2)+mx=3(x-2) 化简,得: (m-1)x=-10 (1)当m=1时,方程无解; (2)若方程(m-1)x=-10的解是x=2,此时原方程也无解,得:m=-4 (3)若方程(m-1)x=-10的解是x=-2,此时原方程也无解,...

x*的 * 是什么意思?

(k+3)/(x-2)=(1-x)/(2-x) k+3=-(1-x)=1+x x=k+3-1=k+2 x-2=0 x=2 k+2=0 k=-2

解: 去分母,等式两边同乘以x-2 ax=4+x-2 (a-1)x=2 a=1时,等式变为0=2,等式恒不成立,a=1满足题意。 a≠1时,x=2/(a-1) 分式方程无解,x=2/(a-1)是分式方程的增根,x=2/(a-1)使得分式方程的分母为0 2/(a-1) -2=0,解得a=2 综上,得:a的值为1...

m/(x²-4) - 1/(x+2) = 0 m/(x²-4) - (x-2) /(x²-4) = 0 (m-x+2) /(x²-4) = 0 {x-(m+2)} / {(x+2)(x-2)} = 0 ∵方程无解 ∴m+2=-2,或m+2=2 ∴m=-4,或m=0

(1)∵a+23=c+56=b4,∴a+2=34b,c+5=32b,∴a=34b-2,c=32b-5,代入2a-b+3c=23得:2(34b-2)-b+3(32b-5)=23,b=425,a=4310,c=385;(2)1x?2+kx+2=4x2?4x+2+k(x?2)(x+2)(x?2)=4x2?4,去分母得:x+2+k(x-2)=4,∵关于x的分式方程1x?2+kx+2...

2/(x-2)+mx/(x²-4)=3/(x+2)两边同乘以x²-4:2(x+2)+mx=3(x-2)2x+4+mx=3x-6(m-1)x=-10x=-10/(m-1)有增根,x=-2或2:x=-2时,-10/(m-1)=-2,m=6x=2时,-10/(m-1)=2,m=-4综上,m=6,或-4

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