Câu 1 : 

a, \(x^3-9x=x\left(x^2-9\right)=x\left(x-3\right)\left(x+3\right)\)

b, \(x^2-25+y^2-2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)

c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)

d, \(\left(x+2\right)^2-x\left(x+3\right)=10\)

\(\Leftrightarrow x^2+4x+4-x^2-3x-10=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

Câu 2 :

a, \(\left(x^2-4y^2\right):\left(x+2y\right)=\left(x-2y\right)\left(x+2y\right):\left(x+2y\right)=x-2y\)

b, \(\frac{-2x}{x-5}-\frac{10}{5-x}=\frac{-2x}{x-5}+\frac{10}{x-5}=\frac{-2x+10}{x-5}=\frac{-2\left(x-5\right)}{x-5}=-2\)

c, \(\frac{3x-3}{x+2}.\frac{x^2-4}{x^2-1}=\frac{3\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}=\frac{3\left(x-2\right)}{x+1}\)

a, Xét tứ giác ABDC có: AM=MD (gt) ; BM=MC (gt)

=> ABDC là hình bình hành

b,Để ABDC là hình thoi => AB = AC => \(\Delta ABC\)cân

c, I đâu ra vậy bạn?

\(a^2-2a+1-b^2\)

\(=\left(a-1\right)^2-b^2\)

\(=\left(a-1-b\right)\left(a-1+b\right)\)

Xét \(\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{1-y}{y^3-1}+\frac{1-x}{x^3-1}=-\frac{1}{x^2+x+1}-\frac{1}{y^2+y+1}\)

\(=-\frac{x^2+y^2+x+y+2}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=-\frac{x^2+y^2+3}{x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1}\)

\(=-\frac{\left(x+y\right)^2-2xy+3}{x^2y^2+x^2+y^2+2xy+2}=-\frac{4-2xy}{x^2y^2+3}=\frac{2\left(xy-2\right)}{x^2y^2+3}\)

từ đó ta có đpcm

\(x\left(x-7\right)-x+7=0\)

\(\Leftrightarrow x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)

Ta có: x(x-7)-(x-7)=0

<=> (x-7)(x-1) = 0 <=> \(\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}}\)