$\frac{1}{x+1}$ – $\frac{2x^{2}-7}{x^{3}+1}$ = $\frac{3}{x^{2}-x+1}$

0 bình luận về “$\frac{1}{x+1}$ – $\frac{2x^{2}-7}{x^{3}+1}$ = $\frac{3}{x^{2}-x+1}$”

1. Giải thích các bước giải:

         $\frac{1}{x+1}$-$\frac{2x²-7}{x³+1}$=$\frac{3}{x²-x+1}$ (x$\neq$-1)

   ⇔ $\frac{1}{x+1}$-$\frac{2x²-7}{(x+1)(x²-x+1)}$=$\frac{3}{x²-x+1}$

   ⇔ $\frac{x²-x+1}{(x+1)(x²-x+1)}$-$\frac{2x²-7}{(x+1)(x²-x+1)}$=$\frac{3(x+1)}{(x+1)(x²-x+1)}$

   ⇒ x²-x+1-2x²+7=3x+3

   ⇔ -x²-4x+5=0

   ⇔ x²+4x-5=0

   ⇔ x²+5x-x-5=0

   ⇔ (x²-x)+(5x-5)=0

   ⇔ x(x-1)+5(x-1)=0 

   ⇔ (x+5)(x-1)=0

   ⇔ \(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=-5(tm)\\x=1(tm)\end{array} \right.\) 

   Vậy S={-5;1}

   Trả lời

2. `1/{x+1}-{2x^2-7}/{x^3+1}=3/{x^2-x+1}`

   ĐKXĐ: `x\ne-1`

   `1/{x+1}-{2x^2-7}/{x^3+1}=3/{x^2-x+1}`

   `⇔1/{x+1}-{2x^2-7}/{(x+1)(x^2-x+1)}=3/{x^2-x+1}`

   `⇔{1(x^2-x+1)}/{(x+1)(x^2-x+1)}-{2x^2-7}/{(x+1)(x^2-x+1)}={3(x+1)}/{(x+1)(x^2-x+1)}`

   `⇒1(x^2-x+1)-(2x^2-7)=3(x+1)`

   `⇔x^2-x+1-2x^2+7=3(x+1)`

   `⇔-x^2-x+8=3x+3`

   `⇔-x^2-x-3x=3-8`

   `⇔-x^2-4x=-5`

   `⇔x^2-4x=5`

   `⇔x^2-4x-5=0`

   `⇔x^2+5x-x-5=0`

   `⇔(x^2+5x)-(x+5)=0`

   `⇔x(x+5)-(x+5)=0`

   `⇔(x+5)(x-1)=0`

   `⇔` \(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\)

   `⇔` \(\left[ \begin{array}{l}x=-5 \text{ (tm ĐKXĐ)}\\x=1\text{ (tm ĐKXĐ)}\end{array} \right.\) 

   Vậy phương trình có tập nghiệm `S={-5;1}`

   Trả lời