`a)` 

Có `{:(BM=9),(AM=6):}}=>(BM)/(AM)=9/6=3/2`

`{:(BN=12),(NC=8):}}=>(BN)/(NC)=12/8=3/2`

Xét `Delta BCA` có \(\dfrac{BM}{AM}=\dfrac{BN}{NC}\left(=\dfrac{3}{2}\right)\)

`=>MN////AC` (Thalè đảo)(dpcm)

`b)`

Xét `Delta BMN` và `Delta BAC` có `MN////AC`

`=>Delta BMN∼Delta BAC`(dpcm)

Có \(\dfrac{BM}{AM}=\dfrac{BN}{NC}\left(cmt\right)\Rightarrow\dfrac{BM}{AM+BM}=\dfrac{BN}{NC+BN}\)

Hay `9/(6+9)=(BN)/(BC)

`=>9/15=(BN)/(BC)`

`=>3/5=(BN)/(BC)`

Có `Delta BMN∼Delta BAC(cmt)` 

`=>` \(\dfrac{BM}{BA}=\dfrac{BN}{BC}=\dfrac{MN}{AC}\)(các cạnh tương ứng)

mà `3/5=(BN)/(BC)`

`=>` $\dfrac{BM}{BA}=\dfrac{BN}{BC}=\dfrac{MN}{AC}=\dfrac{3}{5}$

`a)` 

`M` là trung điểm `BC(g t)=>BM=CM`

Xét `Delta ABM` và `Delta ACM` có:

`{:(AB=AC(g t)),(hat(B)=hat(C)(cmt)),(AM-chung):}}`

`=>Delta ABM=Delta ACM(c.c.c)(dpcm)`

`b)`

Có `Delta ABM=Delta ACM(cmt)`

`=>hat(A_1)=hat(A_2)` ( 2 góc tương ứng )

mà ` AM` nằm trong `AB` và `AC`

Nên `AM` là phân giác `hat(BAC)(dpcm)`

`c)`

Có `MH⊥AB(g t)=>hat(H_1)=90^0`

`MK⊥AC(g t)=>hat(K_1)=90^0`

Xét `Delta AHM` và `Delta AKM` có:

`{:(hat(H_1)=hat(K_1)(=90^0)),(AM-chung),(hat(A_1)=hat(A_2)(cmt)):}}`

`=>Delta AHM=Delta AKM(c.h-g.n)`

`=>MH=MK` ( 2 cạnh tương ứng )(dpcm)

`Delta ABC` có `AB=AC(g t)=>Delta ABC` cân tại `A`

`=>hat(B)=hat(C)(dpcm)`

\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

Vậy `x in {15/4;-3/4}`

\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)

Vậy `x in {+-0,2;2}`