Toán Lớp 8: Tìm x: a, x^2-3x-(x-3)=0 b, 5x(2x-1)-2x+1=0 c, 4x^2-4x+1=0 d, x^4-4x^3+10x^2+40x=0 e, x^3-x-3x^2+3=0

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Toán Lớp 8:  Tìm x: a, x^2-3x-(x-3)=0 b, 5x(2x-1)-2x+1=0 c, 4x^2-4x+1=0 d, x^4-4x^3+10x^2+40x=0 e, x^3-x-3x^2+3=0

haiyemy · Answer

Giải đáp:    Ph&iacute;a dưới   Lời giải và giải thích chi tiết:    a) x^2-3x-(x-3)=0   &hArr;x(x-3)-(x-3)=0   &hArr;(x-1)(x-3)=0   &hArr; ( left[  begin{array}{l}x-1=0  x-3=0 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=1  x=3 end{array}  right. )   Vậy x=1 hoặc x=3   b)5x(2x-1-2x+1=0    &hArr;5x(2x-1)-(2x-1)=0     &hArr;(5x-1)(2x-1)=0     &hArr; ( left[  begin{array}{l}5x-1=0  2x-1=0 end{array}  right. )     &hArr; ( left[  begin{array}{l}x= dfrac{1}{5}  x= dfrac{1}{2} end{array}  right. )   Vậy x=1/5 hoặc x=1/2   c) 4x^2+4x+1=0   &hArr;(2x-1)^2=0   &hArr;2x-1=0   &hArr;x=1/2   Vậy x=1/2   d) x^4-4x^3+10x^2-40x=0 (Đề sai n&ecirc;n chữa lại như thế n&agrave;y)    &hArr;x^3(x-4)+10x(x-4)=0     &hArr;(x^3+10x)(x-4)=0     &hArr;x(x^2+10)(x-4)=0     &hArr;x(x-4)=0( V&igrave; x^2+10 >0 &forall;x)     &hArr; ( left[  begin{array}{l}x=0  x-4=0 end{array}  right. )     &hArr; ( left[  begin{array}{l}x=0  x=4 end{array}  right. )     Vậy x=0 hoặc x=4     e, x^3-x-3x^2+3=0     &hArr;x(x^2-1)-3(x^2-1)=0     &hArr;(x-3)(x^2-1)=0     &hArr;(x-3)(x-1)(x+1)=0     &hArr; ( left[  begin{array}{l}x-3=0  x-1=0  x+1=0 end{array}  right. )      &hArr; ( left[  begin{array}{l}x=3  x=&plusmn;1 end{array}  right. )     Vậy x=3 hoặc x=&plusmn;1

thanoanghha · Answer

Giải đáp+Lời giải và giải thích chi tiết: a)x^2-3x-(x-3)=0 <=>x(x-3)-(x-3)=0 <=>(x-3)(x-1)=0 <=> ( left[ begin{array}{l}x=3 x=1 end{array} right. ) b)5x(2x-1)-2x+1=0 <=>5x(2x-1)-(2x-1)=0 <=>(2x-1)(5x-1)=0 <=> ( left[ begin{array}{l}x= frac{1}{2} x= frac{1}{5} end{array} right. ) c)4x^2-4x+1=0 <=>(2x-1)^2=0 <=>2x-1=0 <=>x=1/2 d)x^4+4x^3+10x^2+40x=0(sai đề nên sửa) <=>x^3(x+4)+10x(x+4)=0 <=>(x+4)(x^3+10x)=0 <=>x(x+4)(x^2+10)=0 Ta thấy x^2+10>=10 ne0 <=>x=-4 hoặc x=0 e)x^3-x-3x^2+3=0 <=>x(x^2-1)-3(x^2-1)=0 <=>(x-1)(x+1)(x-3)=0 <=> ( left[ begin{array}{l}x=±1 x=3 end{array} right. )