Question

даю 65 балов Довести тотожність і знайти значення виразу. ​

Answer 1

$\displaystyle\bf\\\frac{m^{2}+m }{m^{2} +2m+1} -\Big(\frac{1}{m^{2} -1} -\frac{1}{m^{3} -1} \Big)\cdot\Big(\frac{1}{m^{2} } -m\Big)=\\\\\\=\frac{m\cdot(m+1) }{(m+1)^{2} } -\Big(\frac{1}{(m-1)(m+1)} -\frac{1}{(m-1)(m^{2} +m+1)} \Big)\cdot\\\\\\\cdot\Big(\frac{1-m^{3} }{m^{2} } \Big)=\\\\\\=\frac{m}{m+1} -\frac{m^{2}+m+1-m-1 }{(m-1)(m+1)(m^{2} +m+1)} \cdot\frac{1-m^{3} }{m^{2} } =\\\\\\=\frac{m}{m+1} -\frac{m^{2} }{(m+1)(m^{3} -1)} \cdot\frac{1-m^{3} }{m^{2} } =$

$\displaystyle\bf\\=\frac{m}{m+1} +\frac{1}{m+1} =\frac{m+1}{m+1} =1\\\\9)\\\\x^{2} +\frac{1}{x^{2} } =6\\\\\\x-\frac{1}{x} =?\\\\\\\Big(x-\frac{1}{x} \Big)^{2} =x^{2} -\underbrace{2\cdot x\cdot \frac{1}{x} }_{2}+\frac{1}{x^{2} } =x^{2} -2+\frac{1}{x^{2} } =6-2=4\\\\\\x-\frac{1}{x} =\pm \ \sqrt{4} =\pm \ 2$