Hơi xấu =)

Xét $∆MNP$, có:

$MN=NP$ (gt)

$=> \widehat{NMP} = \widehat{NPM}$ $(1)$

Mà $PM$ là tia phân giác (đề bài) nên:

$=> \widehat{NPM} = \widehat{MPQ}$ $(2)$

Từ $(1)(2)$ suy ra $\widehat{NMP} = \widehat{MPQ}$

Mà hai góc so le trong nên $MN // QP$

Xét tứ giác $MNPQ$ có:

$MN//QP$

Suy ra $MNPQ$ là hình thang (đpcm)

Bài \(1\):

\(P=\frac{x+2}{x^2+x+1}-\frac{2}{x-1}-\frac{2x^2+4}{1-x^3}\)

\(P=\frac{x+2}{x^2+x+1}-\frac{2}{x-1}-\frac{2x^2+4}{x^3-1}\)

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

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

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

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

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

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

\(P=\frac{x}{x^2+x+1}\)

Bài \(2\):

\(M=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(M=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{9x^2-4}\)

\(M=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(M=\frac{\left(3x+2\right)-\left(12x-8\right)\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)

\(M=\frac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(M=\frac{\left(3x-12x+3x\right)\left(2+8-6\right)}{\left(3x-2\right)\left(3x+2\right)}\)

\(M=\frac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}\)

\(M=\frac{-2\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}\)

\(M=\frac{-2}{3x+2}\)

Còn mấy bài sau áp dụng y chang bạn tự làm nhé!

Chia mỗi người hai câu cũng hay đó anh.😀