a, Ta có tam giác IMN ~ tam giác ABC (gt) 

\(\dfrac{IM}{AB}=\dfrac{IN}{AC}=\dfrac{MN}{BC}=k=\dfrac{3}{4}\)

b, \(\Rightarrow MN=\dfrac{3BC}{4}=\dfrac{18}{4}=\dfrac{9}{2}cm\)

`Answer:`

a. `(x-2)(2x-1)=5(2-x)`

`<=>(x-2)(2x-1)-5(2-x)=0`

`<=>(x-2)(2x-1)+5(x-2)=0`

`<=>(x-2)(2x-1+5)=0`

`<=>2(x-2)(x+2)=0`

`<=>(x-2)(x+2)=0`

`<=>x-2=0` hoặc `x+2=0`

`<=>x=2` hoặc `x=-2`

b. \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\left(ĐKXĐ:x\ne\pm5\right)\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x^2+25x}{\left(x-5\right)\left(x+5\right)}\)

\(\Rightarrow x^2+10x+25-\left(x-10x+25\right)=x^2+25x\)

\(\Leftrightarrow x^2+10x+25-x+10x-25-x^2-25x=0\)

\(\Leftrightarrow\left(x^2-x^2\right)+\left(10x-x+10x\right)+\left(25-25\right)=0\)

\(\Leftrightarrow19x=0\)

\(\Leftrightarrow x=0\)

bằng 5 nhé 

\(2.\left(x-5\right)\left(x+2\right)=x^2-5x\)

\(\Leftrightarrow2.\left(x^2+2x-5x-10\right)=x^2-5x\)

\(\Leftrightarrow2x^2+4x-10x-20=x^2-5x\)

\(\Leftrightarrow2x^2-x^2+4x-10x+5x-20=0\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow x^2-5x+4x-20=0\)

\(\Leftrightarrow x.\left(x-5\right)+4.\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+4\right).\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=5\end{cases}}}\)

Vậy....

a, Xét tam giác ABC và tam giác OMN có 

^BAC = ^MON = 90⁰

\(\dfrac{AC}{ON}=\dfrac{BC}{MN}=\dfrac{8}{4}=\dfrac{10}{5}=2\)

Vậy tam giác ABC ~ tam giác OMN 

b, \(\dfrac{AB}{OM}=\dfrac{BC}{MN}=\dfrac{AC}{ON}\)( tỉ số đồng dạng )