Toán Lớp 8: Tìm x : 1) x(x – 5)(x + 5) – (x – 2)(x² + 2x + 4) = 3 2) 25x² – 2 = 0 3) (x + 2)² = (2x – 1)²

Question

Toán Lớp 8:  Tìm x : 1) x(x – 5)(x + 5) – (x – 2)(x² + 2x + 4) = 3  2) 25x² – 2 = 0 3) (x + 2)² = (2x – 1)²

cobebongdem · Answer

Giải đáp+Lời giải và giải thích chi tiết:    1) x(x-5)(x+5)-(x-2)(x&sup2;+2x+4)=3   &hArr;x(x&sup2;-25)-(x&sup3;-8)=3   &hArr;x&sup3;-25x-x&sup3;+8=3   &hArr;-25x+8=3   &hArr;-25x=-5   &hArr;x=1/5       Vậy S={1/5}   2) 25x&sup2;-2=0   &hArr;(5x)&sup2;-(&radic;2)&sup2;=0   &hArr;(5x-&radic;2)(5x+&radic;2)=0   $&hArr; left[ begin{matrix} 5x-&radic;2=0   5x+&radic;2=0 end{matrix} right.$   $&hArr; left[ begin{matrix} 5x=&radic;2   5x=-&radic;2 end{matrix} right.$   $&hArr; left[ begin{matrix} x= frac{&radic;2}{5}   x=- frac{&radic;2}{5} end{matrix} right.$        vậy S={ frac{&radic;2}{5};- frac{&radic;2}{5}}   3) (x+2)&sup2;=(2x-1)&sup2;   &hArr;(x+2)&sup2;-(2x-1)&sup2;=0   &hArr;[(x+2)-(2x-1)] [(x+2)+(2x-1)]=0   &hArr;(x+2-2x+1)(x+2+2x-1)=0   &hArr;(-x+3)(3x+1)=0   $&hArr; left[ begin{matrix} -x+3=0   3x+1=0 end{matrix} right.$   $&hArr; left[ begin{matrix} x=3   x=-1/3 end{matrix} right.$        Vậy S={3; -1/3}              Ch&uacute;c bn hc tốt   ????                 $ text{#LinhSad}$

tuyethutanhthu · Answer

1)   $(=) x(x^2-25) - (x^3-8)=3$   $(=) x^3 - 25x - x^3 + 8 = 3$   $(=) -25x + 8 = 3$   $(=) -25x = 3-8$   $(=) -25x = -5$   $=) x=1/5$   2)   $(=) 25x^2=2$   $(=) x^2= 2/25$   $(=) x=  frac{- sqrt2}{5}$ hoặc $x= frac{ sqrt2}{5}$   3)   $(=) x^2 + 4x + 4 - 4x^2 + 4x - 1 =0$   $(=) 3x^2 - 8x - 3 = 0$   $(=) 3x^2 + x - 9x - 3=0$   $(=) x(3x+1) -3(3x+1)=0$   $(=) (x-3) (3x+1) =0$   $(=) x-3=0$ hoặc $3x+1=0$   $=) x=3$ hoặc $x=-1/3$