Đáp án:
Giải thích các bước giải:
`a)`
`(2x-1)^2 = 49`
`⇔ |2x-1| = 7`
`⇔`\(\left[ \begin{array}{l}2x-1=7\\2x-1=-7\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=4\\x=-3\end{array} \right.\)
Vậy `S = {-3,4}`
`b)`
`(5x-3)^2 - (4x-7)^2 = 0`
`⇔ 25x^2 - 30x + 9 - 16x^2 + 56x - 49 = 0`
`⇔ (25x^2-16x^2) + (56x-30x) - (49-9) = 0`
`⇔ 9x^2 + 26x - 40 = 0`
`⇔ (9x^2-10x)+(36x-40)=0`
`⇔ x(9x-10)+4(9x-10)=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 = {-4,10/9}`
`c)`
`9x^2-1=(3x+1)(2x-3)`
`⇔ 9x^2 - 1 = 6x^2 - 7x - 3`
`⇔ 6x^2 - 1 - 9x^2 - 7x - 3 = 0`
`⇔ -3x^2 - 7x - 2 = 0`
`⇔ (-3x^2-x)+(-6x-2)=0`
`⇔ x(3x+1) + 2(3x+1) = 0`
`⇔ (3x+1)(x+2) = 0`
`⇔`\(\left[ \begin{array}{l}3x+1=0\\x+2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac13\\x=-2\end{array} \right.\)
Vậy `S = {-2,-1/3}`
`d)`
`(2x+5)(x-4)=(x-5)(4-x)`
`⇔ 2x^2 - 3x - 20 = 9x - x^2 - 20`
`⇔ 2x^2 - 3x - 20 - 9x + x^2 + 20 = 0`
`⇔ 3x^2 - 12x = 0`
`⇔ 3x(x-4) = 0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\)
Vậy `S = {0,4}`
`e)`
`x^2 - 2x - 8 = 0`
`⇔ (x^2+2x)+(-4x-8)=0`
`⇔ x(x+2)-4(x+2)=0`
`⇔`\(\left[ \begin{array}{l}x-4=0\\x+2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=4\\x=-2\end{array} \right.\)
Vậy `S = {-2,4}`
`f)`
`x^3 - 7x + 6 = 0`
`⇔ x^3 - x^2 + x^2 - x - 6x + 6 = 0`
`⇔ x^2(x-1)+x(x-1)-6(x-1) = 0`
`⇔ (x-1)(x^2+x-6) = 0`
`⇔ (x-1)(x^2+3x-2x-6) = 0`
`⇔ (x-1)[x(x+3)-2(x+3)] = 0`
`⇔ (x-1)(x+3)(x-2) = 0`
`⇔`\(\left[ \begin{array}{l}x-1=0\\x+3=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1\\x=-3\\x=2\end{array} \right.\)
Vậy `S = {-3,1,2}`
