Toán Lớp 7: 1. x^4+x^2+6x-8 2.(x-1)^3+(2x+3)^3=27x^3+8

Question

Toán Lớp 7:  1. x^4+x^2+6x-8 2.(x-1)^3+(2x+3)^3=27x^3+8

ngoclankhue · Answer

#R&ugrave;a     Giải đáp+Lời giải và giải thích chi tiết:      1)x^4+x^2+6x-8=0   &hArr; (x^4-x^3)+(x^3-x^2)+(2x^2-2x)+(8x-8)=0   &hArr; x^3(x-1)+x^2(x-1)+2x(x-1)+8(x-1)=0   &hArr; (x-1)(x^3+x^2+2x+8)=0   &hArr; (x-1)[(x^3+2x^2)-(x^2+2x)+(4x+8)]=0   &hArr; (x-1)[x^2(x+2)-x(x+2)+4(x+2)]=0   &hArr; (x-1)(x+2)(x^2-x+4)=0   V&igrave; x^2-x+4=(x^2-2.x.1/2+1/4)+15/4                     =(x-1/2)^2+1/15>0   &rArr; ( left[  begin{array}{l}x-1=0  x+2=-0 end{array}  right. ) &hArr; ( left[  begin{array}{l}x=1  x=-2 end{array}  right. )    Vậy nghiệm của phương tr&igrave;nh l&agrave;: S={1;-2}   2)(x-1)^3+(2x+3)^3=27x^3+8   &hArr; x^3-3x^2+3x-1+8x^3+36x^2+54x+27-27x^3-8=0   &hArr; -18x^3+33x^2+57x+18=0   &hArr; 6x^3-11x^2-19x-6=0   &hArr; (6x^3+3x^2)-(14x^2+7x)-(12x+6)=0   &hArr; 3x^2(2x+1)-7x(2x+1)-6(2x+1)=0   &hArr; (2x+1)(3x^2-7x-6)=0   &hArr; (2x+1)[(3x^2-9x)+(2x-6)]=0   &hArr; (2x+1)[3x(x-3)+2(x-3)]=0   &hArr; (2x+1)(x-3)(3x+2)=0   &hArr; ( left[  begin{array}{l}2x+1=0  x-3=0  3x+2=0 end{array}  right. ) &hArr; ( left[  begin{array}{l}x= dfrac{-1}{2}  x=3  x= dfrac{-2}{3} end{array}  right. )   Vậy nghiệm của phương tr&igrave;nh l&agrave;: S={-1/2;3;-2/3}

truongankimtrong · Answer

Giải đáp+Lời giải và giải thích chi tiết: 1)x^4+x^2+6x-8=0 <=>x^4-1+x^2+6x-7=0 <=>(x^2-1)(x^2+1)+x^2-x+7x-7=0 <=>(x-1)(x+1)(x^2+1)+x(x-1)+7(x-1)=0 <=>(x-1)(x^3+x+x^2+1+x+7)=0 <=>(x-1)(x^3+x^2+2x+8)=0 <=>(x-1)[x^3+8+x(x+2)]=0 <=>(x-1)[(x+2)(x^2-2x+4)+x(x+2)]=0 <=>(x-1)(x+2)(x^2-x+4)=0 Vì x^2-x+4 =x^2-2*x*1/2+1/4+3 3/4 =(x-1/2)^2+3 3/4>=3 3/4>0AAx <=>(x-1)(x+2)=0 <=>[(x-1=0),(x+2=0):} <=>[(x=1),(x=-2):} Vậy x=1 hoặc x=-2. 2)(x-1)^3+(2x+3)^3=27x^3+8 <=>[(x-1)+(2x+3)][(x-1)^2-(x-1)(2x+3)+(2x+3)^2]=(3x+2)(9x^2-6x+4) <=>(3x+2)(x^2-2x+1-2x^2-x+3+4x^2+12x+9)=(3x+2)(9x^2-6x+4) <=>(3x+2)(3x^2+9x+13)=(3x+2)(9x^2-6x+4) <=>(3x+2)(6x^2-15x-9)=0 <=>(3x+2)(2x^2-5x-3)=0 <=>(3x+2)(2x^2+x-6x-3)=0 <=>(3x+2)[x(2x+1)-3(2x+1)]=0 <=>(3x+2)(2x+1)(x-3)=0 <=> ( left[ begin{array}{l}3x+2=0 2x+1=0 x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=- dfrac23 x=- dfrac12 x=3 end{array} right. ) Vậy S={-2/3,-1/2,3}.