Toán Lớp 8: mọi người giúp mình với ạ tìm x (2x-3)²-4x(x-7)=9 (2x-3)²-9x²=0

Question

Toán Lớp 8:  mọi người giúp mình với ạ tìm x (2x-3)²-4x(x-7)=9 (2x-3)²-9x²=0

ngoclanquynh · Answer

#andy     [ begin{array}{l} a){ left( {2x - 3}  right)^2} - 4x left( {x - 7}  right) = 9     Leftrightarrow 4{x^2} - 12x + 9 - 4{x^2} + 28x = 9     Leftrightarrow 16x + 9 = 9     Leftrightarrow 16x = 0     Leftrightarrow x = 0   b){ left( {2x - 3}  right)^2} - 9{x^2} = 0     Leftrightarrow  left( {2x - 3 - 3x}  right) left( {2x - 3 + 3x}  right) = 0     Leftrightarrow  left( { - x - 3}  right) left( {5x - 3}  right) = 0     Leftrightarrow  left[  begin{array}{l}  - x - 3 = 0   5x - 3 = 0  end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  - 3   x =  dfrac{3}{5}  end{array}  right.     Rightarrow S  in  { 3; dfrac{3}{5} }   end{array} ]

thienthanh · Answer

a) (2x-3)^2 - 4x(x-7) =9 <=> (2x)^2 - 2.2x.3 + 3^2 - 4x^2 + 28x =9 <=> 4x^2 - 12x +9 - 4x^2 + 28x = 9 <=> 16x + 9 = 9 <=> 16x =0 <=> x=0 Vậy x=0 b) (2x-3)^2 - 9x^2=0 <=> (2x-3)^2 - (3x)^2=0 <=> (2x-3-3x)(2x-3+3x)=0 <=> (-x -3)(5x - 3)=0 <=> ( left[ begin{array}{l}-x-3=0 5x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x= dfrac{3}{5} end{array} right. ) Vậy x in { -3 ; 3/5}