a,Xét \(\Delta\) AMC và \(\Delta\) BME có

ME = MA (gt)

BM = MC ( M là trung điểm của BC )

\(\widehat{AMC}=\widehat{BME}\) ( 2 góc đối đỉnh )

=> \(\Delta AMC=\Delta EMB\left(c.g.c\right)\)

=> \(\widehat{ACM=}\widehat{MBE}\) (2 góc tương ứng )

Mà : 2 góc này so le trong

=> AC // BE ( d/h nhận bt 2 đg thẳng // )

b, Xét \(\Delta\) MBK và \(\Delta\) MCE có :

MB = MC (cmt)

BK = EC (gt)

\(\widehat{ACM}=\widehat{MBE}\) (cmt)

=> \(\Delta\) MBK = \(\Delta\) MCE (c.g.c)

=> \(\widehat{BMK}=\widehat{EMC}\) ( 2 góc tương ứng )

Ta có : \(\widehat{BME}+\widehat{EMC}=180^o\) ( kề bù )

Mà : \(\widehat{BMK}=\widehat{EMC}\) ( cmt )

\(\Rightarrow\widehat{BMK}+\widehat{BMK}\) = 180^o

\(\Rightarrow\widehat{KME}=180^o\) => E,M,K thẳng hàng

Ta có : \(2^2+4^2+6^2+...+20^2=\left(1\cdot2\right)^2+\left(2\cdot2\right)^2+\left(2\cdot3\right)^2+...+\left(2\cdot10\right)^2\)

\(=4\cdot1^2+4\cdot2^2+4\cdot3^2+...+4\cdot10^2\)

\(=4\left(1^2+2^2+3^2+...+10^2\right)\)

\(=4\cdot385=1540\)

Ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Rightarrow6x-3y=2x+2y\)

\(\Rightarrow6x-2x=2y+3y\)

\(\Rightarrow4x=5y\Rightarrow\dfrac{x}{y}=\dfrac{5}{4}\)

Vậy ...

\(P=\sqrt{\left(x-\dfrac{3}{4}\right)^2}+\dfrac{1}{4}\)

\(=\left|x-\dfrac{3}{4}\right|+\dfrac{1}{4}\)

Ta có : \(\left|x-\dfrac{3}{4}\right|\ge0\forall x\Rightarrow\left|x-\dfrac{3}{4}\right|+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

\(\Rightarrow P\ge\dfrac{1}{4}\)

Dấu "=" xảy ra

\(\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)

Vậy GTNN của P là \(\dfrac{1}{4}\) khi x = \(\dfrac{3}{4}\)

(x-y)²=4 =>x-y = 2  hoặc x -  y = -2 

có vô số ............

= 370cm                             tick nha

???????????????????????????????????????????????

+) Nếu \(x+y+z\ne0\)

Áp dụng t/c dãy tỉ số = nhau , ta có :

\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)

\(\Rightarrow x+y+z=\dfrac{1}{2}\)

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

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

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

+) Nếu x + y + z = 0

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+z+1}=0\\\dfrac{y}{x+z+1}=0\\\dfrac{z}{x+y-2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

Vậy ...

Vì \(\left(2x-4\right)\left(x+4\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-4< 0\\x+4>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x-4>0\\x+4< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x< 4\\x>-4\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x>4\\x< -4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 2\\x>-4\end{matrix}\right.\\\left\{{}\begin{matrix}x>2\\x< -4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-4< x< 2\\2< x< -4\left(loai\right)\end{matrix}\right.\)

Vậy -4 < x < 2 t/m đề bài