Đáp án:
Giải thích các bước giải:
Đặt
$\sqrt{2x^2+x+9}$ `=m(m>0)`
$\sqrt{2x^2-x+1}$ `=n(n>0)`
`=>m^2=2x^2+x+9`
`=>n^2=2x^2-x+1`
`=>m^2-n^2=2x+8`
`=>m+n=(2x+8)/2 =(m^2-n^2)/2`
`=>(m+n)((m-n)/2 -1)=0`
Lại có `m+n>0`
`=>(m-n)/2 -1=0`
`=>m-n=2`
`=>`$\sqrt{2x^2+x+9}$ `=`$\sqrt{2x^2-x+1}$ `+2`
`=>(\sqrt{2x^2+x+9})^2 =(\sqrt{2x^2-x+1}+2)^2`
`=>2x^2+x+9=2x^2-x+1+4\sqrt{2x^2-x+1}+4`
`<=>2x+4=4\sqrt{2x^2-x+1}`
`<=>x+2=2\sqrt{2x^2-x+1}`
`=>(x+2)^2=(2\sqrt{2x^2-x+1})^2`
`=>x^2+4x+4=4(2x^2-x+1)`
`<=>7x^2-8x=0`
`<=>x(7x-8)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\7x-8=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{8}{7}\end{array} \right.\)
Vậy `S={0,8/7}`
