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x/2x%1+1/x+2=1怎么解

解绝对值不等式就是要分类讨论,去掉绝对值符号 x≥2时,有2x-1+x-2=x+1,x=2 1/2≤x≤2时,有2x-1+2-x=x+1,此时解为1/2≤x≤2 -1≤x≤1/2时,有1-2x+2-x=x+1,x=1/2. x≤-1时,有1-2x+2-x=-x-1,x=2(舍去) 综上,有1/2≤x≤2

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

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)

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...

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

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

两边乘(2x+1)² (x-1)(2x+1)

解答如下:

(2x+1)/(x-3)

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