Toán Lớp 8: giải pt (2x + 7)^2 = 9(x + 2)^2

Question

Toán Lớp 8:  giải pt (2x + 7)^2 = 9(x + 2)^2

tuanh · Answer

$  $    Giải đáp + giải th&iacute;ch c&aacute;c bước giải :    (2x+7)^2=9(x+2)^2   &hArr; (2x+7)^2=3^2(x+2)^2   &hArr; (2x+7)^2=(3x  + 6)^2   &hArr; (2x+7)^2 - (3x+6)^2=0   &hArr; (2x+7-3x-6)(2x + 7 +3x+6)=0   &hArr; (-x + 1) (5x + 13)=0   &hArr;  ( left[  begin{array}{l}-x+1=0  5x+13=0 end{array}  right. ) <=>  ( left[  begin{array}{l}x=1  x= dfrac{-13}{5} end{array}  right. )    Vậy phương tr&igrave;nh c&oacute; tập nghiệm : S={1;(-13)/5}

maianhtuanh · Answer

#andy     [ begin{array}{l}  ;{ left( {2x + 7}  right)^2} ; = 9{ left( {x + 2}  right)^2}     Leftrightarrow { left( {2x + 7}  right)^2} - 9{ left( {x + 2}  right)^2} = 0     Leftrightarrow  left[ { left( {2x + 7}  right) + 3 left( {x + 2}  right)}  right] left[ { left( {2x + 7}  right) - 3 left( {x + 2}  right)}  right] = 0     Leftrightarrow  ; left( {5x{ rm{ }} + 13}  right) left( {1 - x}  right){ rm{ = }}0     Leftrightarrow  left[  begin{array}{l} 5x{ rm{ }} + 13 = 0   1 - x = 0  end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  frac{{ - 13}}{5}   x = 1  end{array}  right.     Rightarrow S  in  { 1; frac{{ - 13}}{5} }   end{array} ]