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x/2x%1+1/x+2=1怎么解

2x/(x-2)-1

方程移项得:x^2+1/x^2+2x+2/x=1 令t=x+1/x, 有:|t|>=1 t^2=x^2+1/x^2+2 方程化为:t^2-2+2t=1 t^2+2t-3=0 (t+3)(t-1)=0 t=-3, t=1(舍去) 所以x+1/x=-3 x^2+3x+1=0 x=(-3+√5)/2, (-3-√5)/2

是不是(1/(x+1)-1/(1-x))? 1/(x+1)-1/(1-x) =1/(x+1)+1/(x-1) =(x-1)/((x+1)(x-1)) + (x+1)/((x+1)(x-1))\ =((x-1)+ (x+1))/((x+1)(x-1)) =2x/((x+1)(x-1)) =2x/(x^2-1)

lim(x→1) (x^2-1)/(2x^2-x-1) =lim(x→1) (x+1)(x-1)/[(2x+1)(x-1)] =lim(x→1) (x+1)/(2x+1) =2/3

2x+1.2x=16 3.2x=16 x=16/3.2 =5

两边乘(2x+1)² (x-1)(2x+1)

解:(2x+1)/(x-1)≤1 [(2x+1)/(x-1)]-[(x-1)/(x-1)]≤0 (2x+1-x+1)/(x-1)≤0 等价转换,两个数的商是非正数,那么积也是非正数。 ∴(x+2)(x-1)≤0 在数轴上表示(序轴标根法)解得 -2≤x≤1 ∵x-1≠0 x≠1 ∴这个不等式的解集为{x|-2≤x<1}

对 九是用换元法。 解: 令1/(x-1)=a 即x=(a+1)/a 那么f[(1/(x-1)]=x/(2x-1)就变为f(a)=[(a+1)/a]/{[2(a+1)/a]-1}=(a+1)/(a+2) 即f(a)=(a+1)/(a+2) 在令a=x+1 所以f(x+1)=(x+1+1)/(x+1+2)=(x+2)/(x+3) 即 f(x+1)=(x+2)/(x+3)

x/(2x-1)+1/(x+2)=1 两边同乘以(2x-1)((x+2) x(x+2)+(2x-1)=(2x-1)(x+2) x²+2x+2x-1=2x²+3x-2 x²-x=1 x²-x+(1/2)²=1+1/4 (x-1/2)²=5/4 x-1/2=(√5)/2或x-1/2=-(√5)/2 即:x=(1+√5)/2或x=(1-√5)/2

解: 1/(x+1)=2/(x-1) 方程两边同时乘(x+1)(x-1) x-1=2(x+1) x-1=2x+2 x-2x=2+1 -x=-3 x=-3 经检验x=-3是原方程的根 2x/(2x+5)+5/(5x-2)=1 方程两边同时乘(2x+5)(5x-2) 2x(5x-2)+5(2x+5)=(2x+5)(5x-2) 10x²-4x+10x+25=10x²+21x-10 -4x...

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