Toán Lớp 8: tìm x (x+2)(x+3)-(x-2)(x+5)=6
(3x+2)(2x+9)-(x+20)(6x+1)=(x+1)-(x-6)
3(2x-1)(3x-1)-(2x-3)(9x-1)=0
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Toán Lớp 8: tìm x (x+2)(x+3)-(x-2)(x+5)=6
(3x+2)(2x+9)-(x+20)(6x+1)=(x+1)-(x-6)
3(2x-1)(3x-1)-(2x-3)(9x-1)=0
Comments ( 1 )
(x+2)(x+3)-(x-2)(x+5)=6
<=>x^2+3x+2x+6-(x^2+5x-2x-10)=6
<=>x^2+5x+6-x^2-5x+2x+10=6
<=>x^2+5x+6-x^2-3x+10=6
<=>(x^2-x^2)+(5x-3x)+(6+10)=6
<=>2x+16=6
<=>2x=6-16
<=>2x=-10
<=>x=-5
Vậy phương trình có nghiệm S={-5}
b)
(3x+2)(2x+9)-(x+20)(6x+1)=(x+1)-(x-6)
<=>6x^2+27x+4x+18-(6x^2+x+120x+20)=x+1-x+6
<=>6x^2+31x+18-(6x^2+121x+20)=7
<=>6x^2+31x+18-6x^2-121x-20=7
<=>(6x^2-6x^2)+(31x-121x)+(18-20)-5=0
<=>-90x-2-7=0
<=>-90x=9
<=>x=-1/10
Vậy phương trình có nghiệm S={-1/10}
c)
3.(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>3.(6x^2-2x-3x+1)-(18x^2-2x-27x+3)=0
<=>3.(6x^2-5x+1)-(18x^2-29x+3)=0
<=>18x^2-15x+3-18x^2+29x-3=0
<=>(18x^2-18x^2)+(-15x+29x)+(3-3)=0
<=>14x=0
<=>x=0
Vậy phương trình có nghiệm S={0}