Question

[tranthuylinh]

giải pt 

\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\)

Answer 1

[Minh Nhân]

\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\left(x\ne0,x\ne1\right)\)

\(\Leftrightarrow\dfrac{140+5x}{x}=\dfrac{150}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(140+5x\right)=150x\)

\(\Leftrightarrow140x+5x^2-140-5x-150x=0\)

\(\Leftrightarrow5x^2-15x-140=0\)

\(\Leftrightarrow x^2-3x-28=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=-4\left(N\right)\end{matrix}\right.\)

\(S=\left\{7,-4\right\}\)

Answer 2

[Trần Ái Linh]

ĐK: `x \ne 0 ; x \ne -1`

`140/x+5=150/(x-1)`

`<=>(140+5x)/x=150/(x-1)`

`<=>(140x+5x)(x-1)=150x`

`<=>5x^2+135x-140=150x`

`<=>5x^2-15x-140=0`

`<=>` \(\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy...