Đáp án:

`m\in {0;{-32}/{15}}`

Giải thích các bước giải:

`\qquad 3x^2-(3m-2)x-(3m+1)=0`

`<=>3x^2+3x-(3m+1)x-(3m+1)=0`

`<=>3x(x+1)-(3m+1)(x+1)=0`

`<=>(x+1)(3x-3m-1)=0`

`<=>`$\left[\begin{array}{l}x+1=0\\3x-3m-1=0\end{array}\right.$`<=>`$\left[\begin{array}{l}x=-1\\x=\dfrac{3m+1}{3}\end{array}\right.$

$\\$

+) `TH1: x_1=-1;x_2={3m+1}/3`

`\qquad 3x_1-5x_2=6`

`<=>3.(-1)-5. {3m+1}/3=6`

`<=>-3.3-15m-5=6.3`

`<=>-15m=32<=>m={-32}/{15}`

$\\$

+) `TH2: x_1={3m+1}/3;x_2=-1`

`\qquad 3x_1-5x_2=6`

`<=>3. {3m+1}/3-5.(-1)=6`

`<=>3m+1+5=6`

`<=>3m=0<=>m=0`

Vậy `m\in {0;{-32}/{15}}` thỏa mãn đề bài 

_______

Cách khác dùng `∆; Viet`

`\qquad 3x^2-(3m-2)x-(3m+1)=0`

`a=3;b=-(3m-2);c=-(3m+1)`

`∆=b^2-4ac=[-(3m-2)]^2-4.3.[-(3m+1)]`

`=9m^2-12m+4+36m+12`

`=9m^2+24m+16=(3m+4)^2\ge 0` với mọi `m`

`=>` Phương trình luôn có hai nghiệm `x_1;x_2` với mọi `m`

Theo hệ thức Viet ta có:

$\quad \begin{cases}x_1+x 2=\dfrac{-b}{a}=\dfrac{3m-2}{3}\\x_1x_2=\dfrac{c}{a}=\dfrac{-(3m+1)}{3}\end{cases}$

Để `3x_1-5x_2=6`

`=>8x_1-5x_1-5x_2=6`

`=>8x_1-5(x_1+x_2)=6`

`=>8x_1=6+5. (3m-2)/3`

`=>8x_1={18+15m-10}/3={15m+8}/3`

$\\$

`\qquad 3x_1-5x_2=6`

`=>3x_1+3x_2-8x_2=6`

`=>3(x_1+x_2)-8x_2=6`

`=>3. {3m-2}/3-6=8x_2`

`=>8x_2=3m-2-6=3m-8`

$\\$

Vì `x_1x_2={-(3m+1)}/3`

`=>8x_1. 8x_2=64. {-(3m+1)}/3`

`=>{15m+8}/3 . (3m-8)={-64}/3 (3m+1)`

`=>(15m+8)(3m-8)=-64(3m+1)`

`=>45m^2-120m+24m-64=-192m-64`

`=>45m^2+96m=0`

`=>3m(15m+32)=0`

`=>`$\left[\begin{array}{l}m=0\\15m+32=0\end{array}\right.$`=>`$\left[\begin{array}{l}m=0\\m=\dfrac{-32}{15}\end{array}\right.$

Vậy `m\in {0;{-32}/{15}}` thỏa mãn đề bài