Lời giải:

a. $\overrightarrow{BC}=(-8; 6)$
Vì đt cần tìm nhận $\overrightarrow{BC}$ là VTPT nên nó có dạng

$-8(x-1)+6(y+2)=0$

$\Leftrightarrow -4x+3y+10=0$

b. Gọi $I(a,b)$ là tâm đường tròn ngoại tiếp tam giác $ABC$. 

Ta có: $IA^2=IB^2=IC^2$

$\Leftrightarrow (a-1)^2+(b+2)^2=(a-5)^2+(b+4)^2=(a+3)^2+(b-2)^2$

\(\Leftrightarrow \left\{\begin{matrix} 8a-4b-36=0\\ -8a+8b-8=0\\ -16a+12b+28=0\end{matrix}\right.\Leftrightarrow a=10; b=11\)

$R^2=IA^2=(a-1)^2+(b+2)^2=(10-1)^2+(11+2)^2=250$

PTĐTr cần tìm là:

$(x-10)^2+(y-11)^2=250$

Lời giải:

a) Xét tam giác $ABM$ và $ACM$ có:

$\widehat{AMB}=\widehat{AMC}=90^0$

$AB=AC$ (do $ABC$ cân tại A)

$AM$ chung

$\Rightarrow \triangle ABM=\triangle ACM$ (ch-cgv)

b) Xét tam giác $ANP$ và $CNM$ có:

$AN=CN$ (do $N$ là trung điểm $AC$)

$NP=NM$ 

$\widehat{ANP}=\widehat{CNM}$ (đối đỉnh)

$\Rightarrow \triangle ANP=\triangle CNM$ (c.g.c)

$\Rightarrow \widehat{APN}=\widehat{CMN}$

Mà 2 góc này ở vị trí so le trong nên $AP\parallel CM$. Mà $AM\perp CM$ nên $AP\perp AM$ (đpcm)

c) 

Từ tam giác bằng nhau phần b suy ra $AP=CM(1)$

Xét tam giác $CMQ$ và $CRQ$ có:

$\widehat{CQM}=\widehat{CQR}=90^0$

$QR=QM$

$QC$ chung

$\Rightarrow \triangle CMQ=\triangle CRQ$ (c.g.c)

$\Rightarrow CM=CR(2)$

Từ $(1);(2)\Rightarrow CR=PA$ (đpcm)

Hình vẽ:

1: \(\left(x+1\right)^2-\left(x-1\right)^2\)

\(=\left(x+1-x+1\right)\left(x+1+x-1\right)\)

\(=2\cdot2x=4x\)

2: \(\left(x+2\right)^2-2\left(x+2\right)\cdot x+x^2\)

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

3: \(\left(x-3\right)^2+\left(x+3\right)^2\)

\(=x^2-6x+9+x^2+6x+9\)

\(=2x^2+18\)

4: \(\left(2x+1\right)\left(2x-1\right)=\left(2x\right)^2-1^2=4x^2-1\)

5: \(\left(y+2\right)^2-\left(y-2\right)^2\)

\(=y^2+4y+4-\left(y^2-4y+4\right)\)

\(=y^2+4y+4-y^2+4y-4=8y\)

6: \(\left(x+4\right)^2-2\left(x+4\right)\left(x-4\right)+\left(x-4\right)^2\)

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

\(=8^2=64\)

7: \(\left(m+1\right)\left(m-1\right)+\left(m+1\right)^2\)

\(=m^2-1+m^2+2m+1\)

\(=2m^2+2m\)

8: \(\left(n+2\right)\left(n-2\right)-\left(n^2-4\right)\)

\(=n^2-4-\left(n^2-4\right)\)

=0

9: \(\left(a+b\right)^2-\left(a-b\right)^2\)

=(a+b+a-b)(a+b-a+b)

=2a*2b=4ab

10: \(\left(p+1\right)^2+\left(p-1\right)^2\)

\(=p^2+2p+1+p^2-2p+1=2p^2+2\)

11: \(\left(x+2\right)^2-2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left(x+2-x+2\right)^2=4^2=16\)

12: \(\left(x-3\right)^2+2\left(x-3\right)\left(x+3\right)+\left(x+3\right)^2\)

\(=\left(x-3+x+3\right)^2=\left(2x\right)^2=4x^2\)

13: \(\left(x-2\right)\cdot\left(x+2\right)\left(x^2+4\right)-\left(x^2-4\right)^2\)

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

\(=\left(x^2-4\right)\left(x^2+4-x^2+4\right)=8\left(x^2-4\right)=8x^2-32\)

14: \(\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)-\left(4x^2-9\right)^2\)

\(=\left(4x^2-9\right)\left(4x^2+9\right)-\left(4x^2-9\right)^2\)

\(=\left(4x^2-9\right)\left(4x^2+9-4x^2+9\right)=18\left(4x^2-9\right)=72x^2-162\)

15: \(\left(4x+1\right)^2-\left(4x-1\right)^2\)

\(=\left(4x+1+4x-1\right)\left(4x+1-4x+1\right)\)

\(=8x\cdot2=16x\)

16: \(\left(x^2+2\right)^2-\left\lbrack\left(x-1\right)\left(x+1\right)\right\rbrack^2\)

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

\(=\left(x^2+2+x^2-1\right)\left(x^2+2-x^2+1\right)=3\left(2x^2+1\right)=6x^2+3\)

17: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)

\(=\left(3x+1-3x+1\right)^2=2^2=4\)

18: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)+\left(x^2-1\right)^2\)

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

\(=\left(x^2-1\right)\cdot\left(x^2+1+x^2-1\right)=2x^2\left(x^2-1\right)=2x^4-2x^2\)

19: \(\left(x^2+3x+2\right)^2-\left(x^2-3x+2\right)^2\)

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

\(=6x\left(2x^2+4\right)=12x^3+24x\)

20: \(\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)-\left(x^2-y^2\right)^2\)

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

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