Toán Lớp 8: tìm x a,4(x+3)(3x-2)-3(x-1)(4x-1)=-27 b,5x(12x+7)-3x(20x-5)=-100 c,0,6x(x-0,5)-0,3x(2x+1,3)=0,138 d,(x+1)(x+2)(x+5)-x ²(x+8)=27

Question

Toán Lớp 8:  tìm x a,4(x+3)(3x-2)-3(x-1)(4x-1)=-27 b,5x(12x+7)-3x(20x-5)=-100 c,0,6x(x-0,5)-0,3x(2x+1,3)=0,138 d,(x+1)(x+2)(x+5)-x ²(x+8)=27

ngoclankhue · Answer

Giải đáp+giải th&iacute;ch c&aacute;c bước giải:   $a,4(x+3)(3x-2)-3(x-1)(4x-1)=-27$   $&hArr;4(3x^2-2x+9x-6)-3(4x^2-x-4x-1)=-27$   $&hArr;4(3x^2+7x-6)-3(4x^2-5x-1)=-27$   $&hArr;12x^2+28x-24-12x^2+15x+3=-27$   $&hArr;43x=-6$   $&hArr;x= dfrac{-6}{43}$   $b,5x(12x+7)-3x(20x-5)=-100$   $&hArr;60x^2+35x-6xx^2+15x=-100$   $&hArr;50x=-100$   $&hArr;x=-2$   $c,0,6x(x-0,5)-0,3x(2x+1,3)=0,138$   $&hArr;0,6x^2-0,3x-0,6x^2-0,39x=0,138$   $&hArr;-0,69x=0,138$   $&hArr;x= dfrac{-1}{5}$   $d,(x+1)(x+2)(x+5)-x &sup2;(x+8)=27$   $&hArr;(x+1)(x^2+5x+2x+10)-x^2(x+8)=27$   $&hArr;(x+1)(x^2+7x+10)-x^2(x+8)=27$   $&hArr;x^3+7x^2+10x+x^2+7x+10-x^3-8x^2=27$   $&hArr;17x=17$   $&hArr;x=1$

truongankimtrong · Answer

Giải đáp + Lời giải và giải thích chi tiết: a) 4(x+3)(3x-2)-3(x-1)(4x-1)=-27 <=>4(3x^2-2x+9x-6)-3(4x^2-x-4x+1)=-27 <=>12x^2-8x+36x-24-12x^2+3x+4x-3=-27 <=>35x-27=-27 <=>35x=0 <=>x=0 Vậy S={0} b) 5x(12x+7)-3x(20x-5)=-100 <=>60x^2+35x-60x^2+15x=-100 <=>50x=-100 <=>x=-2 Vậy S={-2} c) 0,6x(x-0,5)-0,3x(2x+1,3)=0,138 <=>0,6x^2-0,3x-0,6x^2-0,39x=0,138 <=>-0,69x=0,138 <=>x=-(0,138)/(0,69)=-0,2 Vậy S={-0,2} d) (x+1)(x+2)(x+5)-x^2(x+8)=27 <=>(x^2+2x+x+2)(x+5) - x^3 - 8x^2 = 27 <=>x^3+5x^2+2x^2+10x+x^2+5x+2x+10-x^3-8x^2=27 <=>17x+10=27 <=>17x=17 <=>x=1 Vậy S={1}.