Question

[Mina]

giúp mình với ạ. xin cảm ơn<3

Answer 1

[2611]

`x^2+[-18]/[x^2+x]=3-x`         `ĐK: x \ne -1,x \ne 0`

`<=>[x^2(x^2+x)-18]/[x^2+x]=[(3-x)(x^2+x)]/[x^2+x]`

   `=>x^4+x^3-18=3x^2+3x-x^3-x^2`

`<=>x^4+2x^3-2x^2-3x-18=0`

`<=>x^4-2x^3+4x^3-8x^2+6x^2-12x+9x-18=0`

`<=>x^3(x-2)+4x^2(x-2)+6x(x-2)+9(x-2)=0`

`<=>(x-2)(x^3+4x^2+6x+9)=0`

`<=>(x-2)(x^3+3x^2+x^2+3x+3x+9)=0`

`<=>(x-2)[x^2(x+3)+x(x+3)+3(x+3)]=0`

`<=>(x-2)(x+3)(x^2+x+3)=0`

`<=>` $\left[\begin{matrix} x=2 (t/m)\\ x=-3 (t/m)\\x^2+x+3=0\text{ (Vô nghiệm)}\end{matrix}\right.$

Vậy `S={-3;2}`

Answer 2

[Nguyễn Ngọc Huy Toàn]

\(x^2+\dfrac{-18}{x^2+x}=3-x\)

\(\Leftrightarrow x^2-\dfrac{18}{x\left(x+1\right)}=3-x\);\(ĐK:x\ne0;-1\)

\(\Leftrightarrow-\dfrac{18}{x\left(x+1\right)}=3-x-x^2\)

\(\Leftrightarrow\dfrac{18}{x\left(x+1\right)}=x^2+x-3\)

\(\Leftrightarrow\dfrac{18}{x\left(x+1\right)}=x\left(x+1\right)-3\)

Đặt \(x\left(x+1\right)=a\)

\(\Leftrightarrow\dfrac{18}{a}=a-3\)

\(\Leftrightarrow a^2-3a-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-3\end{matrix}\right.\)

Với `x=6`

`=>`\(x^2+x=6\)

`<=>x^2+x-6=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) \((tm)\)

Với `x=-3`

`=>`\(x^2+x=-3\)

`<=>x^2+x+3=0` ( vô lý )

Vậy \(S=\left\{2;-3\right\}\)