Toán Lớp 8: Tìm xbiết: a) 5x ( x + 2) – 3.(x+2) = 0 b) 9????^2 − 25 = 0

Question

Toán Lớp 8:  Tìm xbiết:  a) 5x ( x + 2) – 3.(x+2) = 0  b) 9????^2 − 25 = 0

dananhthumai · Answer

a) 5x(x+2)-3(x+2)=0   &hArr;(x+2).(5x-3)=0.   (nh&acirc;n tử chung : x+2)   $&hArr; left[ begin{matrix} x+2=0  5x-3=0 end{matrix} right.$  $&hArr; left[ begin{matrix} x=-2   x=  dfrac{3}{5} end{matrix} right.$   Vậy x&isin;{-2 ; 3/5}   b)9x^2 - 25=0   &hArr; (3x)^2 - 5^2=0   &hArr;(3x-5).(3x+5)=0   $&hArr; left[ begin{matrix} 3x-5=0  3x+5=0 end{matrix} right.$  $&hArr; left[ begin{matrix} x=  dfrac{5}{3}   x=  dfrac{-5}{3} end{matrix} right.$   Vậy x&isin;{-5/3 ; 5/3}   &Aacute;p dụng HĐT số 3 : a^2 - b^2 =(a-b).(a+b)

huongtructram · Answer

Giải đáp+Lời giải và giải thích chi tiết: a) 5x(x+2) - 3(x+2) = 0 <=> (5x - 3)(x+2) =0 <=> ( left[ begin{array}{l}5x - 3= 0 x+2=0 end{array} right. ) <=> ( left[ begin{array}{l}5x = 3 x=-2 end{array} right. ) <=> ( left[ begin{array}{l}x = dfrac{3}{5} x=-2 end{array} right. ) Vậy x in {3/5; -2} b) 9x^2 - 25 =0 <=> (3x)^2 - 5^2 =0 <=> (3x- 5)(3x +5)=0 <=> ( left[ begin{array}{l}3x - 5=0 3x + 5= 0 end{array} right. ) <=> ( left[ begin{array}{l}3x =5 3x = -5 end{array} right. ) <=> ( left[ begin{array}{l}x = dfrac{5}{3} x = dfrac{-5}{3} end{array} right. ) Vậy x in { +- 5/3}