a: HM là đường trung bình của ΔEBC

=>EH=HB

KM là đường trug bình của ΔFBC

=>FK=KC

ΔAHM có EO//HM

=>AE/AH=AO/AM

ΔAKM có KM//FO

nên AF/AK=AO/AM

=>AE/AH=AF/AK

=>EF//HK

b: ΔAHM có EO//HM

=>MA/MO=HA/HE

=>MA/MO=HA/HB

ΔAKM có FO//KM

=>MA/MO=KA/KF=KA/KC

=>HA/HB=KA/KC

=>HK//BC

=>EF//BC

Answer:

\(x\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)

\(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow x^2+4x+4-\left(x^2-4\right)=0\)

\(\Rightarrow x^2+4x+4-x^2+4=0\)

\(\Rightarrow4x=-8\)

\(\Rightarrow x=-2\)

em lớp 4 nhé nên anh thông cảm

Answer:

Rút gọn phân thức:

\(\frac{4\left(x+1\right)}{\left(3x^2+3x\right)}\)

\(=\frac{4\left(x+1\right)}{3x\left(x+1\right)}\)

\(=\frac{4}{3x}\)

Quy đồng mẫu thức các phân thức sau:

\(\frac{2x}{x^2-1}=\frac{2x^2}{x\left(x-1\right)\left(x+1\right)}\)

\(\frac{7}{x^2-x}=\frac{7\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{7x+7}{x\left(x-1\right)\left(x+1\right)}\)

Tìm x, biết: 

\(\left(x-2\right)\left(2x+1\right)-\left(x-2\right)^2=0\)

\(\Rightarrow\left(x-2\right)\left(2x+1-x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x+1-x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(A=\frac{2x}{x^2-25}+\frac{5}{5-x}-\frac{1}{x+5}\left(ĐKXĐ:x\ne\pm5\right)\)

\(=\frac{2x}{\left(x-5\right)\left(x+5\right)}-\frac{5}{x-5}-\frac{1}{x+5}\)

\(=\frac{2x}{\left(x-5\right)\left(x+5\right)}-\frac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=-\frac{4}{x-5}\)