To solve the equation x, 2 / 2 of the square minus 1 / 1 of x = 1 do somebody a favour

To solve the equation x, 2 / 2 of the square minus 1 / 1 of x = 1 do somebody a favour

2/(x+1)(x-1)-1/(x-1)=1
Multiply by (x + 1) (x-1)
2-(x+1)=(x+1)(x-1)
x²+x-2=0
(x+2)(x-1)=0
x=-2,x=1
The fractional equation needs to be tested
The common denominator is 0 when x = 1
It's zenggen, it's renqu
So x = - 2

The inverse of X squared plus 2x plus the sum of 15 divided by 5 minus X

(-x²+2x+15)/(5-x)
=(x²-2x-15)/(x-5)
=(x-5)(x+3)/(x-5)
=x+3

If the solution of equation 2x minus a equals 1 is a positive number, the value range of a is obtained

2x-a = 1, then x = (a + 1) / 2 > 0
Only a + 1 > 0, a > - 1
Is that clear?

The root of the equation 2x ^ 2-mx + 1 / 2 = 0 of X is greater than or equal to - 1, less than or equal to 2, and the value range of M is obtained

-1=2,m=0
m