Tìm x `1) 5x ( x – 3 ) – 2x = 6 = 0` `2) x ³ – 8 = (x – 2 ) ³`

2 bình luận về “Tìm x `1) 5x ( x – 3 ) – 2x = 6 = 0` `2) x ³ – 8 = (x – 2 ) ³`”

1. 1)

   5x.(x-3)-2x+6=0

   =>5x.(x-3)-2.(x-3)=0

   =>(5x-2).(x-3)=0

   =>[(5x-2=0),(x-3=0):}

   =>[(5x=2),(x=3):}

   =>[(x=2/5),(x=3):}

   Vậy x in {2/5;3}

   2)

   x^3-8=(x-2)^3

   =>x^3-2^3=(x-2)^3

   =>(x-2).(x^2-2x+4)=(x-2)^3

   =>(x-2).(x^2-2x+4)-(x-2)^3=0

   =>(x-2).[x^2-2x+4-(x-2)^2]=0

   =>(x-2).[x^2-2x+4-(x^2-4x+4)]=0

   =>(x-2).(x^2-2x+4-x^2+4x-4)=0

   =>(x-2).2x=0

   =>[(x-2=0),(2x=0):}

   =>[(x=2),(x=0):}

   Vậy x in {2;0}

   Trả lời
2. Giải đáp:

    1)

   5x.(x-3)-2x+6=0

   => 5x.(x-3)-2.(x-3)=0

   => (5x-2).(x-3)=0

   => 5x-2=0 hoặc x-3=0

   => 5x=2 hoặc x=3

   => x=\frac{2}{5} hoặc x=3

   Vậy x\in{\frac{2}{5};3}

   2)

   x^{3}-8=(x-2)^{3}

   => (x^{3}-2^{3})-(x-2)^{3}=0

   => (x-2).(x^{2}+2x+4)-(x-2).(x-2)^{2}=0

   => (x-2).[x^{2}+2x+4-(x^{2}-4x+4)]=0

   => (x-2).(x^{2}+2x+4-x^{2}+4x-4)=0

   => (x-2).6x=0

   =>x-2=0 hoặc 6x=0

   => x=2 hoặc x=0

   Vậy x\in{2;0}

   #DYNA

    

   Trả lời