Đáp án:
Giải thích các bước giải:
`sqrt{3x+1}-sqrt{6-x}+3x^2-14x-8=0`
`ĐKXĐ:-1/3<=x<=6`
`<=>sqrt{3x+1}-4+1-sqrt{6-x}+3x^2-14x-5=0`
`<=>(3x+1-16)/(sqrt{3x+1}+4)+(1-6+x)/(1+sqrt{6-x})+3x^2-15x+x-5=0`
`<=>(3x-15)/(sqrt{3x+1}+4)+(x-5)/(1+sqrt{6-x})+3x(x-5)+(x-5)=0`
`<=>(3(x-5))/(sqrt{3x+1}+4)+(x-5)/(1+sqrt{6-x})+(x-5)(3x+1)=0`
`<=>(x-5)(3/(sqrt{3x+1}+4)+1/(1+sqrt{6-x})+3x+1)=0`
Vì `3/(sqrt{3x+1}),1/(1+sqrt{6-x})>0`
`->x-5=0->x=5`
