Решите пожалуйста, даю 20 баллов. Created by iana180305. algebra-ru

Решите пожалуйста, даю 20 баллов...

LizaOper

Ответ:

$\left\Х\begin{array}{l}\dfrac{1}{x+1}+\dfrac{1}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{3}{y-2}=1 \end{array}\right\ \ a=\dfrac{1}{x+1}\ ,\ b=\dfrac{1}{y-2}\ \ \ \left\{\begin{array}{l}a+b=5\\5a-3b=1\end{array}\right\\\\\\\left\{\begin{array}{l}b=5-a\\5a-3(5-a)=1\end{array}\right\ \ \left\{\begin{array}{l}b=5-a\\8a-15=1\end{array}\right\ \ \left\{\begin{array}{l}b=3\\a=2\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{1}{x+1}=2 \\\ \dfrac{1}{y-2}=3\end{array}\right$

$\left\{\begin{array}{l}x+1=\dfrac{1}{2} \\y-2=\dfrac{1}{3} \end{array}\right\ \ \left\{\begin{array}{l}x=-\dfrac{1}{2}\\\ \ \ y=2\dfrac{1}{3}\end{array}\right\ \ \ Otvet:\ \ \Big(-\dfrac{1}{2}\ ;\ 2\dfrac{1}{3}\, \Big)\ .$

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NNNLLL54

Verified answer

Ответ:

1/(x+1) + 1/(y-2)= 5

5/(x+1)-3/(y-2)=1

У нас есть похожие части, введем новые переменные

1/(х+1) обозначим буквой a

1/(y-2) - буквой b, тогда

$\left \{ {{a+b=5} \atop {5a-3b=1}} \right. \left \{ {{a=5-b} \atop {5(5-b)-3b=1}} \right. \left \{ {{a=5-b} \atop {2-5b-3b=1}} \right. \left \{ {{a=5-b} \atop {-8b=-24}} \right. \left \{ {{a=5-3} \atop {b=3}} \right.\left \{ {{a=2} \atop {b=3}} \right.$ ⇒

1/(x+1)=2

1=2x+2

2x=-1

x=-0.5

1/(y-2)=3

1=3y-6

3y=7

y=2 1/3

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