Toán Lớp 8: Tìm x 1, x(1-2x)+(x-2)(2x-3)=0 2, (x-1)(x+2)-x-2=0 3, x^3-16x=0 4, (2x-3)^2+(x-3)(3-2x)=0

Question

Toán Lớp 8:  Tìm x 1, x(1-2x)+(x-2)(2x-3)=0 2, (x-1)(x+2)-x-2=0 3, x^3-16x=0 4, (2x-3)^2+(x-3)(3-2x)=0

trucoanh55 · Answer

Giải đáp:   1)S={1}   2) S={&plusmn;2}   3) S={0;&plusmn;4}   4)S={0;3/2}   Lời giải và giải thích chi tiết:    &hArr;1, x(1-2x)+(x-2)(2x-3)=0     &hArr;x-2x^2 + 2x^2-3x-4x+6=0     &hArr;6-6x=0     &hArr;x=1     Vậy S={1}     2, (x-1)(x+2)-x-2=0     (x-1)(x+2) - (x+2) =0     (x+2) (x-1-1)=0     (x+2)(x-2)=0     &hArr; ( left[  begin{array}{l}x+2=0  x-2=0 end{array}  right. )     &hArr; ( left[  begin{array}{l}x=-2  x=2 end{array}  right. )     Vậy S={&plusmn;2}     3, x^3-16x=0     &hArr;x(x^2-16)=0     &hArr;x(x-4)(x+4)=0     &hArr; ( left[  begin{array}{l}x=0  x-4=0  x+4=0 end{array}  right. )     &hArr; ( left[  begin{array}{l}x=0  x=&plusmn;4 end{array}  right. )     Vậy S={0;&plusmn;4}     4, (2x-3)^2+(x-3)(3-2x)=0     &hArr;(2x-3)^2 - (x-3)(2x-3)=0     &hArr;(2x-3)(2x-3-x+3)=0     &hArr;x(2x-3)=0     &hArr; ( left[  begin{array}{l}x=0  2x-3=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=0  x= dfrac{3}{2}  end{array}  right. )    Vậy S={0;3/2}

lanngocha · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    1)    $x(1-2x)+(x-2)(2x-3)=0$     $x - 2x&sup2; + 2x&sup2; - 7x + 6 = 0$     $-6x + 6 = 0$     $6x = 6$     $x = 1$     2)     $(x-1)(x+2) - x - 2 = 0$     $(x-1)(x+2) - (x+2) = 0$     $(x+2)(x-1-1) = 0$     $(x+2)(x-2) = 0$     $x&sup2; - 4 = 0$     $x&sup2; = 4$     $x = &plusmn;2$     3, $x&sup3;-16x=0$     $x(x&sup2; - 16) = 0$      ( left[  begin{array}{l}x=0  x&sup2;=16 end{array}  right. )      &rArr;  ( left[  begin{array}{l}x=0  x=&plusmn;4 end{array}  right. )      4, $(2x-3)&sup2;+(x-3)(3-2x)=0$     $(2x - 3)&sup2; - (x-3)(2x-3) = 0$     $(2x-3)(2x - 3 -x + 3) = 0$     $(2x - 3)x = 0$     ( left[  begin{array}{l}x=0  x= dfrac{3}{2} end{array}  right. )