Đáp án:

 `1)`

`5x^2-6x+1=0`

`<=>5x^2-5x-x+1=0`

`<=>5x.(x-1)-(x-1)=0`

`<=>(5x-1)(x-1)=0`

`<=>`\(\left[ \begin{array}{l}5x-1=0\\x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=1\end{array} \right.\) 

 Vậy `S={1/5;1}`

`2)`

`2x^2-4x-5=0`

`<=>2x^2-4x+2-7=0`

`<=>(\sqrt{2}.x)^2-2.\sqrt{2}x.\sqrt{2}+(\sqrt{2})^2-7=0`

`<=>(\sqrt{2}.x-\sqrt{2})^2=7`

`<=>(\sqrt{2})^2.(x-1)^2=7`

`<=>2(x-1)^2=7`

`<=>(x-1)^2=(7)/(2)=(±(\sqrt{14})/(2))^2`

`<=>`\(\left[ \begin{array}{l}x-1=\dfrac{\sqrt{14}}{2}\\x-1=-\dfrac{\sqrt{14}}{2}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{14}}{2}+1\\x=-\dfrac{\sqrt{14}}{2}+1\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{2+\sqrt{14}}{2}\\x=\dfrac{2-\sqrt{14}}{2}\end{array} \right.\) 

Vậy `S={(2-\sqrt{14})/(2);(2+\sqrt{14})/(2)}`

`3)`

`x^2+2x-3=0`

`<=>x^2+3x-x-3=0`

`<=>x.(x+3)-(x+3)=0`

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

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

`<=>`\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\) 

Vậy `S={-3;1}`

`4)`

`x^2-2x-8=0`

`<=>x^2-4x+2x-8=0`

`<=>x.(x-4)+2.(x-4)=0`

`<=>(x+2).(x-4)=0`

`<=>`\(\left[ \begin{array}{l}x+2=0\\x-4=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-2\\x=4\end{array} \right.\) 

Vậy `S={-2;4}`

`5)`

`x^2-15x-16=0`

`<=>x^2-16x+x-16=0`

`<=>x.(x-16)+(x-16)=0`

`<=>(x+1).(x-16)=0`

`<=>`\(\left[ \begin{array}{l}x+1=0\\x-16=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-1\\x=16\end{array} \right.\) 

Vậy `S={-1;16}`

Đáp án:

`1)S={1;1/5}`

`2)S={\frac{\sqrt{7}+\sqrt{2}}{\sqrt{2}} ; \frac{\sqrt{2}-\sqrt{7}}{\sqrt{2}}}`

`3)S={-3;1}`

`4)S={4;-2}`

`5)S={16;-1}`

Giải thích các bước giải:

`1)5x²-6x+1=0`

`⇔5x²-5x-x+1=0`

`⇔(5x²-5x)-(x-1)=0`

`⇔5x(x-1)-(x-1)=0`

`⇔(x-1)(5x-1)=0`

`(1)x-1=0⇔x=1`

`(2)5x-1=0⇔x=1/5`

Vậy `S={1;1/5}`

`2)2x²-4x-5=0`

`⇔2x²-4x+2-7=0`
`⇔(`$\sqrt[]{2}$`x)^2-2.`$\sqrt[]{2}$`x.`$\sqrt[]{2}$`+(`$\sqrt[]{2}$`)^2-7=0`
`⇔(`$\sqrt[]{2}$`x-`$\sqrt[]{2}$`)^2-7=0`
`⇔(`$\sqrt[]{2}$`x-`$\sqrt[]{2}$`)^2=7`
`⇔2(x^2-2x+1)=7`
`⇔2(x-1)^2=7`
`⇔(x-1)^2=7/2`
`⇔(x-1)^2=(±(\sqrt{7})/(\sqrt{2}))^2`
`⇔`\(\left[ \begin{array}{l}x-1=\dfrac{\sqrt{7}}{\sqrt{2}}\\x-1=-\dfrac{\sqrt{7}}{\sqrt{2}}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{7}}{\sqrt{2}}+1\\x=-\dfrac{\sqrt{7}}{\sqrt{2}}+1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{2}}\\x=\dfrac{\sqrt{2}-\sqrt{7}}{\sqrt{2}}\end{array} \right.\)

Vậy `S={\frac{\sqrt{7}+\sqrt{2}}{\sqrt{2}} ; \frac{\sqrt{2}-\sqrt{7}}{\sqrt{2}}}`

`3)x²+2x-3=0`

`⇔x²+3x-x-3=0`

`⇔(x²+3x)-(x+3)=0`

`⇔x(x+3)-(x+3)=0`

`⇔(x+3)(x-1)=0`

`(1)x+3=0⇔x=-3`

`(2)x-1=0⇔x=1`

Vậy `S={-3;1}`

`4)x²-2x-8=0`

`⇔x²-4x+2x-8=0`

`⇔(x²-4x)+(2x-8)=0`

`⇔x(x-4)+2(x-4)=0`

`⇔(x-4)(x+2)=0`

`(1)x-4=0⇔x=4`

`(2)x+2=0⇔x=-2`

Vậy `S={4;-2}`

`5)x²-15x-16=0`

`⇔x²-16x+x-16=0`

`⇔(x²-16x)+(x-16)=0`

`⇔x(x-16)+(x-16)=0`

`⇔(x-16)(x+1)=0`

`(1)x-16=0⇔x=16`

`(2)x+1=0⇔x=-1`

Vậy `S={16;-1}`