Đáp án `+` Giải thích các bước giải `!`
`7)`
`5x(x-3)-2x+6 = 0`
`<=> 5x(x-3)-(2x-6) = 0`
`<=> 5x(x-3)-2(x-3) = 0`
`<=> (5x-2)(x-3) = 0`
`<=>` \(\left[ \begin{array}{l}5x-2=0\\x-3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}5x=2\\x=3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{2}{5}\\x=3\end{array} \right.\)
Vậy `S= {2/5; 3}`
`8)`
`9(3x-2) = x^2(2-3x)`
`<=> 9(3x-2)+x^2(3x-2) = 0`
`<=> (9+x^2)(3x-2) = 0`
`<=>` \(\left[ \begin{array}{l}9+x^2=0\\3x-2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x^2=-9\\3x=2\end{array} \right.\)
Vì `x^2 >= 0` `AA x`
Mà `x^2 =-9` (vô lý)
`<=> x = 2/3`
Vậy `S= { 2/3}`
`9)`
`x^2(x-2)+14 = 7x`
`<=> x^3-2x^2-7x+14 = 0`
`<=> (x^3-2x^2)-(7x-14) = 0`
`<=> x^2(x-2)-7(x-2) = 0`
`<=> (x^2-7)(x-2) = 0`
`<=>` \(\left[ \begin{array}{l}x^2-7=0\\x-2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x^2=7\\x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\sqrt{7}\\x=-\sqrt{7}\\x=2\end{array} \right.\)
Vậy `S= {\sqrt{7}; -\sqrt{7}; 2}`
`h)`
`(2x+5)(2x-7)-(2x-3)^2 = 36`
`<=> (4x^2-14x+10x-35)-(4x^2-12x+9) = 36`
`<=> 4x^2-4x-35-4x^2+12x-9 = 36`
`<=> (4x^2-4x^2)+(-4x+12x)+(-35-9) = 36`
`<=> 8x-44 = 36`
`<=> 8x = 80`
`<=> x = 10`
Vậy `S= {10}`
