3c+9÷(c-3)...

Egorkalyadov03 @Egorkalyadov03

July 2022 1 6 Report

Выполните действия
b^3-8/b^2-9×b+3/b^2+2b+4
c^2+6c+9/c^3+27×c^2-3c+9/3c+9÷(c-3)

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гюйс

Verified answer

$\frac{ b^{3}-8 }{ b^{2} -9} * \frac{b+3}{ b^{2} +2b+4} = \frac{(b-2)(b^{2} +b+4)}{(b-3)(b+3)} * \frac{b+3}{(b+2)(b+2)} = \frac{ b^{3}-8}{(b-3)(b+2)(b+2)} \\ \\ \\ \frac{ c^{2} +6c+9}{ c^{3} +27} * \frac{ c^{2} -3c+9}{3c+9}:(c-3)= \frac{(c+3)(c+3)}{(c+3)( c^{2} -3c+9)} * \frac{c^{2} -3c+9}{3(c+3)} * \frac{1}{c-3} = \frac{1}{3(c-3)}$

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