Question

\(\dfrac{1}{2x^2-x+1}+\dfrac{3}{2x^2-x+3}=\dfrac{10}{2x^2-x+7}\)

Answer 1

Máy bài này khá dễ chỉ cần suy nghĩ tí là làm được.

Đặt `2x^2-x=a(a>=-1/8)`

`pt<=>1/(a+1)+3/(a+3)=10/(a+7)`

`<=>(a+3)(a+7)+3(a+1)(a+7)=10(a+1)(a+3)`

`<=>a^2+10a+21+3(a^2+8a+7)=10(a^2+4a+3)`

`<=>a^2+3a^2+10a+24a+21+21=10a^2+40a+30`

`<=>4a^2+34a+42=10a^2+40a+30`

`<=>6a^2+6a-12=0`

`<=>a^2+a-2=0`

`a+b+c=0`

`=>a_1=1,a_2=-2(l)`

`a=1=>2x^2-x=1`

`=>2x^2-x-1=0`

`a+b+c=0`

`=>x_1=1,x_1=-1/2`

Vậy `S={1,-1/2}`

Answer 2

Đặt \(2x^2-x+1=a\left(a\ge\dfrac{7}{8}\right)\)

PTTT : \(\dfrac{1}{a}+\dfrac{3}{a+2}=\dfrac{10}{a+6}\)

\(\Leftrightarrow\left(a+2\right)\left(a+6\right)+3a\left(a+6\right)=10a\left(a+2\right)\)

\(\Leftrightarrow a^2+2a+6a+12+3a^2+18a=10a^2+20a\)

\(\Leftrightarrow-6a^2+6a+12=0\)

\(\Leftrightarrow\left(a+1\right)\left(a-2\right)=0\)

\(\Leftrightarrow a=2\)

-Thay lại a = 2 ta được : \(2x^2-x-1=0\)

<=> \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...