Mister Exam

Lang:

How to use it?
How do you in partial fractions?:
(((y-y)/(3*y-3))+(1/(y-1)))/(y+1)/3
(a^7*a)^2/(-a)^15
-1+x/(x^2-2*x+2)+(x^2+3*x+2)/(x^2-x+2)
45*x*y^2/(75*y^2)
Factor polynomial:
c^2-9
x^2-25
e^x+5/x
z^3+z-10
Least common denominator:
((z^2+1)^2/(4*z^2))/(13+12*((z^2+1)/2*z*z))
-z/1-z/2
y/(y+3)+y/3+2/y
y/(y-3)+3/(3-y)
Factor squared:
-y^4+8*y^2+9
-y^4+8*y^2-4
-y^4-y^2-2
y^4+8*y^2-4
Identical expressions
x^(two *n)/(x^n- one)^ two
x to the power of (2 multiply by n) divide by (x to the power of n minus 1) squared
x to the power of (two multiply by n) divide by (x to the power of n minus one) to the power of two
x(2*n)/(xn-1)2
x2*n/xn-12
x^(2*n)/(x^n-1)²
x to the power of (2*n)/(x to the power of n-1) to the power of 2
x^(2n)/(x^n-1)^2
x(2n)/(xn-1)2
x2n/xn-12
x^2n/x^n-1^2
x^(2*n) divide by (x^n-1)^2
Similar expressions
x^(2*n)/(x^n+1)^2

How do you x^(2*n)/(x^n-1)^2 in partial fractions?

You have entered [src]

    2*n  
   x     
---------
        2
/ n    \ 
\x  - 1/

$$\frac{x^{2 n}}{\left(x^{n} - 1\right)^{2}}$$

x^(2*n)/(x^n - 1)^2

Fraction decomposition [src]

1 + (-1 + x^n)^(-2) + 2/(-1 + x^n)

$$1 + \frac{2}{x^{n} - 1} + \frac{1}{\left(x^{n} - 1\right)^{2}}$$

        1           2   
1 + ---------- + -------
             2         n
    /      n\    -1 + x 
    \-1 + x /

Common denominator [src]

               n   
       -1 + 2*x    
1 + ---------------
         2*n      n
    1 + x    - 2*x

$$\frac{2 x^{n} - 1}{x^{2 n} - 2 x^{n} + 1} + 1$$

1 + (-1 + 2*x^n)/(1 + x^(2*n) - 2*x^n)

Powers [src]

      n   
  / 2\    
  \x /    
----------
         2
/      n\ 
\-1 + x /

$$\frac{\left(x^{2}\right)^{n}}{\left(x^{n} - 1\right)^{2}}$$

(x^2)^n/(-1 + x^n)^2

Numerical answer [src]

x^(2.0*n)/(-1.0 + x^n)^2

x^(2.0*n)/(-1.0 + x^n)^2