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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.

化简即得2/x=3-k,x=2/(3-k),显然当k=3时分母为0,分式无意义,此时x无解

解:去分母得k(x-1)+(x+k)(x+1)=(x+1)(x-1) kx-k+x²+kx+x+k=x²-1 (2k+1)x=-1 x=-1/(2k+1) 因为x为负数,所以-1/(2k+1)-1/2

3+(2-x)/(x-3)=k/(3-x) [3(x-3)+(2-x)]/(x-3)=-k/(x-3) (3x-9+2-x)/(x-3)=-k/(x-3) (2x-7)/(x-3)=-k/(x-3) 2x-7=-k k=7-2x 方程无解时,x-3=0 x=3 代入k=7-2x =7-2×3 =1 因此,k=1时,方程无解。

1/(x-2)+k/(x+2)=4/(x²-4); (x+2+kx-2k)/(x²-4)=4/(x²-4); x+2+kx-2k=4; (k+1)x=2k+2; x=2; 增根2吧,k∈R 如果本题有什么不明白可以追问,如果满意记得采纳 如果有其他问题请采纳本题后另发点击向我求助,答题不易,请谅解,谢...

题目写的好一点 这样看不懂

方程两边同时乘以(x+2)(x-2)可得:x(x-2)-(x+2)2=k,∴-4-6x=k,则:x=?4?k6.又∵原方程无解,故x可能取值为2或-2,∴①当x=2时,k=-16;②当x=-2时,k=8.故满足条件的k值可能为-16或8.

(x + k)/(x + 1) - k/(x - 1) = 1 (x-1)(x + k)/(x + 1)(x - 1) - k(x + 1)/(x + 1)(x - 1) = 1 (x - 1)(x + k) - k(x + 1) = (x + 1)(x - 1) x^2 + kx - x - k - kx - k = x^2 - 1 -x - 2k = -1 x = 1 - 2k x = 1 - 2k 这即是原分式方程的解, ...

x-2分之2+x分之x-2+x(x-2)分之2x+k=0x-1+1-2+2/(x-2)+k=0x-2+k+2/(x-2)=0x(x-2)-(2-k)(x-2)+2=0x²-2x-(2x-4-kx+2k)+2=0x²-2x-2x+4+kx-2k+2=0x²-2x-2x+kx-2k+2+4=0x²-4x+kx-2k+6=0x²-(4-k)x-2(k-3)=0只有一个解,即判别...

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