bài 4
`a)`
`(x-1)(x+2)-(x+3)^2=5`
`<=>x^2+2x-x-2-(x^2+2.3.x+3^2)-5=0`
`<=>x^2+x-2-x^2-6x-9-5=0`
`<=>(x^2-x^2)+(x-6x)+(-2-9-5)=0`
`<=>-5x-16=0`
`<=>-5x=16`
`<=>x=-16/5`
vậy `S={-16/5}`
`b)`
`(2x-1)^2-2(x+3)(x-3)=-1`
`<=>(2x)^2-2.2x.1+1^2-2(x^2-3^2)+1=0`
`<=>4x^2-4x+1-2x^2+18+1=0`
`<=>(4x^2-2x^2)-4x+(18+1+1)=0`
`<=>2x^2-4x+20=0`
`<=>2x^2-4x+2+18=0`
`<=>2(x^2-2x+1)+18=0`
`<=>2(x-1)^2+18=0`
`<=>2(x-1)^2=-18`
`<=>(x-1)^2=-9 ( `vô lí vì `(x-1)^2≥0∀x)`
vậy pt vô nghiệm
