Toán Lớp 8: bài 2 tìm x a)4x^2-6x=0 b) (2x+5)^2=(x-9)^2 c) x^2-7x+12=0

Question

Toán Lớp 8:  bài 2 tìm x a)4x^2-6x=0 b) (2x+5)^2=(x-9)^2 c) x^2-7x+12=0

thanhtung · Answer

$ text{4x&sup2;-6x=0}$   $ text{&hArr;2x(2x-3)=0}$   $ text{&hArr; ( left[  begin{array}{l}2x=0  2x-3=0 end{array}  right. ) }$   $ text{&hArr; ( left[  begin{array}{l}x=0  x= dfrac{3}{2} end{array}  right. ) }$   _________________________________________________________   $ text{(2x+5)&sup2;=(x-9)&sup2;}$   $ text{&hArr;(2x+5)&sup2;-(x-9)&sup2;=0}$   $ text{&hArr;(2x+5-x+9)(2x+5+x-9)=0}$   $ text{&hArr;(x+14)(3x-4)=0}$   $ text{&hArr; ( left[  begin{array}{l}x+14=0  3x-4=0 end{array}  right. )}$   $ text{&hArr; ( left[  begin{array}{l}x=-14  x= dfrac{4}{3} end{array}  right. )}$   __________________________________________________________   $ text{x&sup2;-7x+12=0}$   $ text{&hArr;x&sup2;-3x-4x+12=0}$   $ text{&hArr;(x&sup2;-3x)-(4x-12)=0}$   $ text{&hArr;x(x-3)-4(x-3)}$   $ text{&hArr;(x-3)(x-4)=0}$   $ text{&hArr; ( left[  begin{array}{l}x-3=0  x-4=0 end{array}  right. ) }$   $ text{&hArr; ( left[  begin{array}{l}x=3  x=4 end{array}  right. ) }$

ngochuong2k2 · Answer

Giải đáp:     C&acirc;u a)  H&igrave;nh b&ecirc;n dưới nka     C&acirc;u b) :    (     2    x    +    5     )^2 &minus;     (    x   &minus;  9   )^2         =       (     2    x    +    5    +    x    &minus;    9     )     .     (     2    x    +    5    &minus;    x    +    9     )      (2x+5+x&minus;9).(2x+5&minus;x+9)      =       (     3    x    &minus;    4     )     .     (     x    +    14     )               C&acirc;u c)  H&igrave;nh b&ecirc;n dưới nka