Toán Lớp 8: CAu 1 tìm x biết a) x(x-3)-2x+6=0 b,(x+3)^2-(x+2)(x-2)=4x=17

Question

Toán Lớp 8:  CAu 1 tìm x biết a)  x(x-3)-2x+6=0 b,(x+3)^2-(x+2)(x-2)=4x=17

quynhthanhnghi · Answer

$ textit{x( x - 3 ) - 2x + 6 = 0}$   $ textit{&rArr;x( x - 3 ) - 2(x - 3 ) = 0}$   $ textit{&rArr;( x - 3 )(x - 2 ) = 0}$   $ textit{&rArr;$ left[ begin{matrix} x - 3 = 0   x - 2 = 0 end{matrix} right.$&hArr;$ left[ begin{matrix} x = 3   x = 2 end{matrix} right.$}$   $ text{b, ( x + 3 $)^{2}$ - ( x + 2 )( x - 2 ) -4x = 17}$   $ text{&rArr; $x^{2}$ + 6x + 9 - ( $x^{2}$ - 4 ) - 4x = 17}$   $ text{&rArr;$x^{2}$ + 6x + 9 - $x^{2}$ +4 - 4x =17}$    $ text{&rArr;2x + 13 = 17}$   $ text{&rArr;2x = 4}$   $ text{&rArr; x = 2}$

danvanthu · Answer

$a) x(x-3)-2x+6=0$   $&hArr;x(x-3)-(2x-6)=0$   $&hArr;x(x-3)-2(x-3)=0$   $&hArr;(x-2)(x-3)=0$   $&hArr;$ ( left[  begin{array}{l}x-2=0  x-3=0 end{array}  right. )   $&hArr;$ ( left[  begin{array}{l}x=2  x=3 end{array}  right. )   $b,(x+3)^2-(x+2)(x-2)-4x=17$   $&hArr;(x^2+6x+9)-(x^2-4)-4x=17$   $&hArr;x^2+6x+9-x^2+4-4x=17$   $&hArr;2x+13=17$   $&hArr;2x=4$   $&hArr;x=2$