Toán Lớp 8: giải các phương trình sau a, 3-4x(25-2x)=8x^2+x-300 b, x- 2x-5/5 + x+8/6=7 + x-1/3 c, 2x(x-3)+5(x-3)=0 c, (x^2-4)-(x-2)(3-2x)=0 mình cầ

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Toán Lớp 8:  giải các phương trình sau a, 3-4x(25-2x)=8x^2+x-300 b, x- 2x-5/5 + x+8/6=7 + x-1/3 c, 2x(x-3)+5(x-3)=0 c, (x^2-4)-(x-2)(3-2x)=0 mình cần gấp

phuongthaongoc · Answer

a)3-4x(25-2x)=8x&sup2;+x-300   &hArr;3-100x+8x&sup2;=8x&sup2;+x-300   &hArr;-100x+8x&sup2;-8x&sup2;-x=-300-3   &hArr;-101x=-303   &hArr;x=(-303):(-101)   &hArr;x=3   Vậy S={3}   b)x-(2x-5)/5+(x+8)/6=7+(x-1)/3   &hArr;(30x)/30-[6(2x-5)]/30+[5(x+8)]/30=210/30+[10(x-1)]/30   &rArr;30x-6(2x-5)+5(x+8)=210+10(x-1)   &hArr;30x-12x+30+5x+40=210+10x-10   &hArr;23x+70=10x+200   &hArr;23x-10x=200-70   &hArr;13x=130   &hArr;x=130:13   &hArr;x=10   Vậy S={10}   c)2x(x-3)+5(x-3)=0   &hArr;(x-3)(2x+5)=0   (1)x-3=0&hArr;x=3   (2)2x+5=0&hArr;x=-5/2   Vậy S={3;-5/2}   d)(x&sup2;-4)-(x-2)(3-2x)=0   &hArr;(x+2)(x-2)-(x-2)(3-2x)=0   &hArr;(x-2)[(x+2)-(3-2x)]=0   &hArr;(x-2)(x+2-3+2x)=0   &hArr;(x-2)(3x-1)=0   (1)x-2=0&hArr;x=2   (2)3x-1=0&hArr;x=1/3   Vậy S={2;1/3}

tuyen98 · Answer

a) 3-4x(25-2x)=8x^2+x-300 <=>3-100x+8x^2=8x^2+x-300 <=>8x^2-8x^2-100x-x=-300-3 <=>-101x=-303 <=>x=3 Vậy S={3} $ $ b) x-(2x-5)/5+(x+8)/6=7+(x-1)/3 <=>(30x-6(2x-5)+5(x+8))/30=(7.30+10(x-1))/30 <=>30x-12x+30+5x+40=210+10x-10 <=>23x+70=10x+200 <=>23x-10x=200-70 <=>13x=130 <=>x=10 Vậy S={10} $ $ c) 2x(x-3)+5(x-3)=0 <=>(2x+5)(x-3)=0 <=> ( left[ begin{array}{l}2x+5=0 x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=- dfrac{5}{2} x=3 end{array} right. ) Vậy S={-5/2;3} $ $ d) (x^2-4)-(x-2)(3-2x)=0 <=>(x-2)(x+2)-(x-2)(3-2x)=0 <=>(x-2)(x+2-3+2x)=0 <=>(x-2)(3x-1)=0 <=> ( left[ begin{array}{l}x-2=0 3x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=2 x= dfrac{1}{3} end{array} right. ) Vậy S={1/3;2}