срочно решите уравнение. Created by danilqr525. algebra-ru

срочно решите уравнение...

table123

$1) \ \dfrac{2m^{2} -12}{m^{2} -4} +\dfrac{5-2m}{m-2}= \dfrac{2m^{2} -12}{(m-2)(m+2)} +\dfrac{5-2m}{m-2}= \\\\\\=\dfrac{2m^{2}-12+(5-2m)(m+2) }{(m-2)(m+2)}= \dfrac{2m^{2} -12+5m+10-2m^{2} -4m}{(m-2)(m+2)}=\\\\\\=\dfrac{m-2}{(m-2)(m+2)}=\boxed{\dfrac{1}{m+2}}$

$2) \ \dfrac{3n-2}{n+3} +\dfrac{9-6n-3n^{2} }{n^{2} +6n+9}=\dfrac{3n-2}{n+3} +\dfrac{9-6n-3n^{2} }{(n+3)^{2} }=\\\\\\=\dfrac{(3n-2)(n+3)+9-6n-3n^{2} }{(n+3)^{2} }=\dfrac{3n^{2}+9n-2n-6+9-6n-3n^{2} }{(n+3)^{2} }=\\\\\\=\dfrac{n+3}{(n+3)^{2} }=\boxed{\dfrac{1}{n+3}}$

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danilqr525 можно ещё второе пожалуйста?

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Ответ:

Объяснение:

1) =(2m^2 -12)/(m-2)(m+2) + (5-2m)/(m-2) =(2m^2-12+(m+2)*(5-2m)/(m-2)(m+2)=

(2m^2-12+5m-2m^2+10-4m)/(m-2)(m+2)=(m-2)/(m-2)(m+2) =1/(m+2)

( ^ -знак степени)

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danilqr525 огромное спасибо

Answers & Comments

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3 use the phrases from ex 2 to complete the sentences

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