bước 4 quên đổi dấu ạ

\(a,\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\left(ĐKXĐ:x\ne3;x\ne-1\right)\\ \dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}\\ \dfrac{1}{\left(3-x\right)\left(x+1\right)}-\dfrac{3-x}{\left(3-x\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{\left(3-x\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(3-x\right)\left(x+1\right)}\\ 1-3+x=-x^2-x+x^2-2x+1\\ x-2=1-3x\\ x+3x=1+2\\ 4x=3\\ x=\dfrac{3}{4}\left(T/m\right)\)

Vậy phương trình có nghiệm `x=3/4`

\(d,\dfrac{7x^2-14x-5}{15}=\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}\\ \dfrac{7x^2-14x-5}{15}=\dfrac{3\left(4x^2+4x+1\right)}{15}-\dfrac{5\left(x^2-2x+1\right)}{15}\\ \dfrac{7x^2-14x-5}{15}=\dfrac{12x^2+12x+3-5x^2+10x-5}{15}\\ 7x^2-14x-5=7x^2+22x-2\\ -14x-22x=-2+5\\ -36x=3\\ x=-\dfrac{1}{12}\)

`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)`

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