What is (b-1)/(b+2-\frac{b^2+2){b-(b-2)/(b+1)}} ?

The solution to (b-1)/(b+2-\frac{b^2+2){b-(b-2)/(b+1)}} is b-1

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simplify (b-1)/(b+2-\frac{b^2+2){b-(b-2)/(b+1)}}

simplify

\frac{b - 1}{b + 2 - \frac{b ^{2} + 2}{b - \frac{b - 2}{b + 1}}}

b - 1

Solution steps

\frac{b - 1}{b + 2 - \frac{b ^{2} + 2}{b - \frac{b - 2}{b + 1}}}

Simplify

b + 2 - \frac{b ^{2} + 2}{b - \frac{b - 2}{b + 1}} : 1

= \frac{b - 1}{1}

Apply the fraction rule:

\frac{a}{1} = a

= b - 1

3 (3 x)

4 (x + 1)^{2}

(a + b) + (a - b)

\frac{x ^{3} + x ^{2} - 10 x + 8}{x - 2}

\frac{5}{3} - \frac{2}{3}

What is (b-1)/(b+2-\frac{b^2+2){b-(b-2)/(b+1)}} ?
The solution to (b-1)/(b+2-\frac{b^2+2){b-(b-2)/(b+1)}} is b-1