bạn ơi điểm E là điểm nào vậy? M A B I O F D C K

ĐKXĐ: \(x\ne-1;y\ne3\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+1}=u\\\dfrac{1}{y-3}=v\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}u-2v=-\dfrac{1}{2}\\3u+v=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u-2v=-\dfrac{1}{2}\\6u+2v=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u-2v=-\dfrac{1}{2}\\7u=\dfrac{7}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{1}{2}\\v=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{2}\\\dfrac{1}{y-3}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+1=2\\y-3=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=a\left(a\ne0\right)\\y-3=b\left(b\ne0\right)\end{matrix}\right.\)

Hệ pt trở thành \(\left\{{}\begin{matrix}\dfrac{1}{a}-\dfrac{2}{b}=-\dfrac{1}{2}\\\dfrac{3}{a}+\dfrac{1}{b}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}-\dfrac{2}{b}=-\dfrac{1}{2}\\\dfrac{6}{a}+\dfrac{2}{b}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{a}=\dfrac{7}{2}\\\dfrac{1}{a}-\dfrac{2}{b}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\left(tm\right)\\\dfrac{1}{2}-\dfrac{2}{b}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\left(tm\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2\\y-3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Vậy x=1;y=5

có tam giác ABC đều (gt)

=> góc A = góc B = góc C (đn) (1)

     AB = AC = BC 

AB = BM + MA

AC = AN + NC

BC = BE + CE

mà BE = CN = AM (gt)    (2)

=> BM = AN = CE                     (3)

(1)(2)(3) => tam giác AMN = tam giác CNE = tam giác BEM (c - g - c)

=> MN = NE = EM

=> tam giác MEN đều

a, R' ≥ 2 ≤ 6

b, R' = 2 cm

c, R' = 1 cm

 Mong giúp ích cho bn !

x=10

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