Môn Toán Lớp 8: (3x+2)(2x^2-7x+15)=(2x-1)(3x+2)(x-3) tìm x

2 bình luận về “Môn Toán Lớp 8: (3x+2)(2x^2-7x+15)=(2x-1)(3x+2)(x-3) tìm x”

1.    

   

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2. Answer:

    (3x+2).(2x^{2}-7x+15)=(2x-1).(3x+2).(x-3)

   => 3x.(2x^{2}-7x+15)+2.(2x^{2}-7x+15)=(6x^{2}+4x-3x-2).(x-3)

   => 6x^{3}-21x^{2}+45x+4x^{2}-14x+30=(6x^{2}+x-2).(x-3)

   => 6x^{3}-17x^{2}+31x+30=6x^{2}.(x-3)+x.(x-3)-2.(x-3)

   => 6x^{3}-17x^{2}+31x+30=6x^{3}-18x^{2}+x^{2}-3x-2x+6

   => 6x^{3}-17x^{2}+31x+30=6x^{3}-17x^{2}-5x+6

   => 6x^{3}-17x^{2}+31x+30-6x^{3}+17x^{2}+5x-6=0

   => (6x^{3}-6x^{3})+(17x^{2}-17x^{2})+(31x+5x)+(30-6)=0

   => 36x+24=0

   => 36x=-24

   => x=-\frac{24}{36}

   => x=-\frac{2}{3}

   Vậy x=-\frac{2}{3}

    

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