Toán Lớp 8: Tìm x: $a,3x(x-2)-x+2=0$ $b,x^2(x+1)+2x(x+1)=0$ $c,x^4(x-2)-2+x=0$ $d,x(2x-3)-2(3-2x)=0$ $e,5(x+3)=2x(3+x)$ $f,(x-2)(x^2+2x+5)+2(x-2)(x

Question

Toán Lớp 8:  Tìm x: $a,3x(x-2)-x+2=0$ $b,x^2(x+1)+2x(x+1)=0$ $c,x^4(x-2)-2+x=0$ $d,x(2x-3)-2(3-2x)=0$ $e,5(x+3)=2x(3+x)$ $f,(x-2)(x^2+2x+5)+2(x-2)(x+2)-5(x-2)=0$

hienchaudiem · Answer

$ $ a, 3x(x-2)-x+2=0 <=>3x(x-2)-(x-2)=0 <=>(x-2)(3x-1)=0 TH1 : x-2=0 <=>x=2 TH2 : 3x-1=0 <=>x=1/3 Vậy x=2,x=1/3 b, x^2(x+1)+2x(x+1)=0 <=> (x+1)(x^2+2x)=0 <=> x(x+1)(x+2)=0 TH1 : x=0 TH2 : x+1=0 <=>x=-1 TH3 : x+2=0 <=>x=-2 Vậy x=0,x=-1,x=-2 c, x^4 (x-2)-2+x=0 <=>x^4 (x-2) - (2-x)=0 <=>x^4(x-2)+(x-2)=0 <=>(x-2)(x^4+1)=0 <=>x-2=0 (Do x^4+1>= 1 > 0∀x) <=>x=2 Vậy x=2 d, x(2x-3)-2(3-2x)=0 <=>x(2x-3)+2(2x-3)=0 <=>(2x-3)(x+2)=0 TH1 : 2x-3=0 <=>x=3/2 TH2 : x+2=0 <=>x=-2 Vậy x=3/2 ,x=-2 e, 5(x+3)=2x(3+x) <=>5(x+3)-2x(x+3)=0 <=>(x+3)(5-2x)=0 TH1 : x+3=0 <=>x=-3 TH2 : 5-2x=0 <=>x=5/2 vậy x=-3,x=5/2 f, (x-2)(x^2+2x+5)+2(x-2)(x+2) - 5 (x-2)=0 <=> (x-2)(x^2+2x+5 + 2x+4 - 5)=0 <=>(x-2)(x^2 + 4x + 4)=0 <=> (x-2)(x+2)^2=0 TH1 : x-2=0 <=>x=2 TH2 : (x+2)^2=0 <=>x+2=0 <=>x=-2 Vậy x=2,x=-2

thienthanh · Answer

Giải đáp +Lời giải và giải thích chi tiết:     $a) 3x( x - 2 ) - x + 2 = 0$   $&hArr; ( x - 2 )( 3x - 1 ) = 0$   &hArr; $ left[ begin{matrix} x = 2    x =  dfrac{1}{3}  end{matrix} right.$   $b) x^2( x + 1 ) + 2x( x + 1 ) = 0$   $&hArr; x( x + 2 )( x + 1 ) = 0$   &hArr; $ left[ begin{matrix} x = 0   x = -2    x = -1 end{matrix} right.$   $c) x^4( x - 2 ) - 2 + x = 0$   $&hArr; x^4( x - 2 ) + ( x - 2 ) = 0$   $&hArr; ( x^4 + 1 )( x - 2 ) = 0$   $&hArr; x - 2  = 0 ( x^4 + 1 > 0 với &forall;x )$   $&hArr; x = 2$   $d) x( 2x - 3 ) - 2( 3 - 2x ) = 0$   $&hArr; x( 2x - 3 ) + 2( 2x - 3 ) = 0$   $&hArr; ( x + 2 )( 2x - 3 ) = 0$   &hArr; $ left[ begin{matrix} x = -2    x =  dfrac{3}{2}  end{matrix} right.$   $e) 5( x + 3 ) = 2x( 3 + x )$   $&hArr; 5( x + 3 ) - 2x( x + 3 ) = 0$   $&hArr; ( 5 - 2x )( x + 3 ) = 0$   &hArr; $ left[ begin{matrix} x =  dfrac{5}{2}    x = -3 end{matrix} right.$   $f) ( x - 2 )( x^2 + 2x + 5 ) + 2( x - 2 )( x + 2 ) - 5( x - 2 ) = 0$   $&hArr; ( x - 2 )( x^2 + 2x + 5 + 2x + 4 - 5 ) = 0$   $&hArr; ( x - 2 )( x^2 + 4x + 4 ) = 0$   $&hArr; ( x - 2 )( x + 2 )^2 = 0$   &hArr; $ left[ begin{matrix} x - 2 = 0   x + 2 = 0 end{matrix} right.$   &hArr; $ left[ begin{matrix} x = 2    x = -2  end{matrix} right.$