`b)(5x-3)²-(4x-7)²=0`
`⇔(5x-3+4x-7)(5x-3-4x+7)=0`
`⇔(9x-10)(x+4)=0`
`⇔`\(\left[ \begin{array}{l}9x-10=0\\x+4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{10}{9}\\x=-4\end{array} \right.\)
Vậy `S={10/9;-4}`
`d)(x+2)²=9(x²-4x+4)`
`⇔(x+2)²=9(x-2)²`
`⇔(x+2)²=[3(x-2)]²`
`⇔(x+2)²-[3(x-2)]²=0`
`⇔[x+2+3(x-2)][x+2-3(x-2)]=0`
`⇔(x+2+3x-6)(x+2-3x+6)=0`
`⇔(4x-4)(-2x+8)=0`
`⇔`\(\left[ \begin{array}{l}4x-4=0\\-2x+8=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}4x=4\\-2x=-8\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1\\x=4\end{array} \right.\)
Vậy `S={1;4}`
`f)(5x²-2x+10)²=(3x²+10x-8)²`
`⇔(5x²-2x+10)²-(3x²+10x-8)²=0`
`⇔(5x²-2x+10+3x²+10x-8)[5x²-2x+10-(3x²+10x-8)]=0`
`⇔(8x²+8x+2)(5x²-2x+10-3x²-10x+8)=0`
`⇔2(4x²+4x+1)(2x²-12x+18)=0`
`⇔4(2x+1)²(x²-6x+9)=0`
`⇔4(2x+1)²(x-3)²=0`
`⇔(2x+1)²(x-3)²=0`
`⇔`\(\left[ \begin{array}{l}(2x+1)²=0\\(x-3)²=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x+1=0\\x-3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=3\end{array} \right.\)
Vậy `S={-1/2;3}`
