Tìm X a)(X-2)^2-(X+3).(X-3)=1 b)(X+2)^3-X^2.(X+6)=4 c)(X+5)^2-(X+2).(X-3)=4 d)(X+3)^2-(X+4).(X-1)=7 e)(X-1)^2+(2X+1)^2-3.(X-1

2 bình luận về “Tìm X a)(X-2)^2-(X+3).(X-3)=1 b)(X+2)^3-X^2.(X+6)=4 c)(X+5)^2-(X+2).(X-3)=4 d)(X+3)^2-(X+4).(X-1)=7 e)(X-1)^2+(2X+1)^2-3.(X-1”

1. a)

   (x-2)^2-(x+3).(x-3)=1

   =>x^2-4x+4-(x^2-3^2)=1

   =>x^2-4x+4-x^2+9=1

   =>(x^2-x^2)-4x=1-4-9

   =>-4.x=-12

   =>x=(-12):(-4)

   =>x=3

   Vậy x=3

   b)

   (x+2)^3-x^2.(x+6)=4

   =>x^3+6x^2+12x+8-x^3-6x^2=4

   =>(x^3-x^3)+(6x^2-6x^2)+12.x=4-8

   =>12.x=-4

   =>x=-1/3

   Vậy x=-1/3

   c)

   (x+5)^2-(x+2).(x-3)=4

   =>x^2+10x+25-(x^2-3x+2x-6)=4

   =>x^2+10x+25-(x^2-x-6)=4

   =>x^2+10x+25-x^2+x+6=4

   =>(x^2-x^2)+(10x+x)=4-25-6

   =>11.x=-27

   =>x=-27/11

   Vậy x=-27/11

   d)

   (x+3)^2-(x+4).(x-1)=7

   =>x^2+6x+9-(x^2-x+4x-4)=7

   =>x^2+6x+9-(x^2+3x-4)=7

   =>x^2+6x+9-x^2-3x+4=7

   =>(x^2-x^2)+(6x-3x)=7-9-4

   =>3.x=-6

   =>x=-2

   Vậy x=-2

   e)

   (x-1)^2+(2x+1)^2-3.(x-1).(1+x)=5

   =>x^2-2x+1+4x^2+4x+1-3.(x-1).(x+1)=5

   =>x^2-2x+1+4x^2+4x+1-3.(x^2-1)=5

   =>x^2-2x+1+4x^2+4x+1-3x^2+3=5

   =>(x^2+4x^2-3x^2)+(-2x+4x)+(1+1+3-5)=0

   =>2x^2+2x=0

   =>2x.(x+1)=0

   =>[(2x=0),(x+1=0):}

   =>[(x=0),(x=-1):}

   Vậy x in {0;-1}

   Trả lời
2. a) (x-2)^2-(x+3)(x-3)=1

   =>x^2-4x+4-x^2+9=1

   =>(x^2-x^2)-4x+(4+9)=1

   =>-4x+13=1

   =>-4x=1-13

   =>-4x=-12

   =>x=-12:(-4)

   =>x=3

   Vậy x=3

   b) (x+2)^3-x^2(x+6)=4

   =>x^3+6x^2+12x+8-x^3-6x^2=4

   =>(x^3-x^3)+(6x^2-6x^2)+12x+8=4

   =>12x=4-8

   =>12x=-4

   =>x=-4:12

   =>x=(-1)/3

   Vậy x=(-1)/3

   c) (x+5)^2-(x+2)(x-3)=4

   =>x^2+10x+25-(x^2-3x+2x-6)=4

   =>x^2+10x+25-x^2+3x-2x+6=4

   =>(x^2-x^2)+(10x+3x-2x)+(25+6)=4

   =>11x+31=4

   =>11x=4-31

   =>11x=-27

   =>x=(-27)/11

   Vậy x=(-27)/11

   d) (x+3)^2-(x+4)(x-1)=7

   =>x^2+6x+9-(x^2-x+4x-4)=7

   =>x^2+6x+9-x^2+x-4x+4=7

   =>(x^2-x^2)+(6x+x-4x)+(9+4)=7

   =>3x+13=7

   =>3x=7-13

   =>3x=-6

   =>x=-6:3

   =>x=-2

   Vậy x=-2

   e) (x-1)^2+(2x+1)^2-3(x-1)(1+x)=5

   =>x^2-2x+1+4x^2+4x+1-3(x^2-1)=5

   =>x^2-2x+1+4x^2+4x+1-3x^2+3=5

   =>(x^2+4x^2-3x^2)-(2x-4x)+(1+1+3)=5

   =>2x^2+2x+5=5

   =>2x^2+2x+5-5=0

   =>2x^2+2x=0

   =>2x(x+1)=0

   => $[\begin{array}{l}2x=0\\ x+1=0\end{array}$ 

   => $[\begin{array}{l}x=0\\ x=-1\end{array}$ 

   Vậy x\in{0;-1}

    

   Trả lời