Toán Lớp 8: tìm x nha 4, (x+3) ³-(3x+10)x+(2x+1)(4x ²-2x+1)=28 5, x ²+2x+1=0 6, x+5x ² =0 7, x+1=(x+1) ² 8, x ² +x=0 9, x ³ -0,25x=0

Question

Toán Lớp 8:  tìm x nha 4, (x+3) ³-(3x+10)x+(2x+1)(4x ²-2x+1)=28 5, x ²+2x+1=0 6, x+5x ² =0 7, x+1=(x+1) ² 8, x ² +x=0 9, x ³ -0,25x=0

mocmien87 · Answer

4)   (x+3)^3- (3x+10)x + (2x+1)(4x^2-2x+1) = 28   => (x^3 + 3 . x^2 . 3 + 3 . x . 3^2 + 3^3) - (3x . x + 10.x) + (2x+1)[(2x)^2 - 2x . 1 + 1^2] = 28   => (x^3 + 9x^2 + 27x + 27) - (3x^2 + 10x)  + [ (2x)^3 + 1^3]=28   => (x^3 + 9x^2 + 27x+27)-(3x^2 + 10x) + (8x^3 + 1) =28   =>x^3+9x^2 + 27x+27-3x^2 - 10x + 8x^3  +1 = 28   => (x^3 + 8x^3) + (9x^2 - 3x^2)  +(27x - 10x) + (27x + 1) = 28   => 9x^3 + 6x^2 + 17x + 28=28   => 9x^3 + 6x^2 + 17x= 0   => x (9x^2 + 6x + 17) =0   =>x=0 hoặc 9x^2+ 6x+17=0   +)x=0   +)9x^2 +6x+17=0   => (9x^2 + 6x+1) + 16=0   => [ (3x)^2  + 2 . 3x . 1 + 1^2]+16=0   =>(3x+1)^2+16=0 (kh&ocirc;ng xảy ra)   Vậy x=0   5)   x^2+2x+1=0   =>x^2+2.x.1+1^2=0   =>(x+1)^2=0   =>x+1=0   =>x=-1   Vậy x=-1   6)   x + 5x^2=0   => x (1+5x)=0   =>x=0 hoặc 1+5x=0   +)x=0   +)1+5x=0=>5x=-1=>x=-1/5   Vậy x in{0;-1/5}   7)   (x+1)=(x+1)^2   =>(x+1)^2-(x+1)=0   =>(x+1) [ (x+1) - 1]=0   => (x+1)(x+1-1)=0   => x (x+1)=0   =>x=0 hoặc x+1=0   +)x=0   +)x+1=0=>x=-1   Vậy x in{0;-1}   8)   x^2+x=0   =>x(x+1)=0   =>x=0 hoặc x+1=0   +)x=0   +)x+1=0=>x=-1   Vậy x in{0;-1}   9)   x^3-0,25x=0   =>x(x^2-0,25)=0   =>x [ x^2 - (0,5)^2]=0   =>x (x-0,5)(x+0,5)=0   =>x=0 hoặc x-0,5=0 hoặc x+0,5=0   +)x=0   +)x-0,5=0=>x=0,5   +)x+0,5=0=>x=-0,5   Vậy x in{0;0,5 ; -0,5}

buianhtuan · Answer

Giải đáp + Lời giải và giải thích chi tiết:   4) (x+3)^3 - (3x+10) x + (2x+1) (4x^2-2x+1) = 28   &hArr; x^3 + 6x^2 + 9x + 3x^2 + 18x + 27 - 3x^2 - x + 8x^3 - 4x^2 + 2x + 4x^2 - 2x + 1 = 28   &hArr; 9x^3 + 6x^2 + 26x + 28 = 28   &hArr; 9x^3 + 6x^2 + 26x = 0   &hArr; x (9x^2+6x+26) = 0   +) 9x^2 + 6x + 26 = 0 (L)   +) x = 0   Vậy x = 0   5) x^2 + 2x + 1 = 0   &hArr; (x+1)^2 = 0   &hArr; x + 1 = 0   &hArr; x = -1   Vậy x = -1   6) x + 5x^2 = 0   &hArr; x (1+5x) = 0   +) x = 0   +) 1 + 5x = 0   &hArr; 5x = -1   &hArr;  x = -1/5   Vậy x = 0 hoặc x = -1/5   7) x + 1 = (x+1)^2   &hArr; x + 1 = x^2 + 2x + 1    &hArr; x - x^2 - 2x = 1 - 1   &hArr; -x^2 - x = 0   &hArr; -x (x-1) = 0   +) -x = 0   &hArr; x = 0   +) x - 1 = 0   &hArr; x = 1   Vậy x = 0  hoặc x = 1   8) x^2 + x = 0   &hArr; x (x+1) = 0   +) x = 0   +) x + 1 = 0   &hArr; x = -1   Vậy x = 0 hoặc x = -1   9) x^3 - 0,25x = 0   &hArr; x (x^2-0,25) = 0   +) x = 0   +) x^2 - 0,25 = 0   &hArr; x^2 = 0,25   &hArr; x = +- 0,5   Vậy x = 0 hoặc x = +- 0,5