tìm x: x(2x-4)=(2x+3)(x+2)-39 (3x-5)^2=(x+1)^2 x^3-8=(x-2)(x+4)(x+10) làm giúp pls

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1. Giải thích các bước giải:

   `x(2x-4)=(2x+3)(x+2)-39`

   `=>2x^2-4x=2x^2+4x+3x+6-39`

   `=>2x^2-4x=2x^2+7x-33`

   `=>11x-33=0`

   `=>11(x-3)=0`

   `=>x-3=0=>x=3`

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

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

   `=>(3x-5-x-1)(3x-5+x+1)=0`

   `=>(2x-6)(4x-4)=0`

   `=>2(x-3).4(x-1)=0`

   `=>`\(\left[ \begin{array}{l}x-3=0\\x-1=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)

   `x^3-8=(x-2)(x+4)(x+10)`

   `=>(x-2)(x^2+2x+4)-(x-2)(x^2+14x+40)=0`

   `=>(x-2)(x^2+2x+4-x^2-14x-40)=0`

   `=>(x-2)(-12x-36)=0`

   `=>-12(x-2)(x+3)=0`

   `=>`\(\left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)

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2. a,x(2x-4)=(2x+3)(x+2)-39

   ⇔2x²-4x=2x²+4x+3x+6-39

   ⇔-11x=-33

   ⇔x=3

   Vậy x=3

   b,(3x-5)²=(x+1)² 

   ⇔(3x-5)²-(x+1)²=0

   ⇔(3x-5-x-1)(3x-5+x+1)=0

   ⇔(2x-6)(4x-4)=0

   ⇔8(x-3)(x-1)=0

   ⇔\(\left[ \begin{array}{l}x-3=0\\x-1=0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)

   Vậy x=3;x=1

   c,x³-8=(x-2)(x+4)(x+10)

   ⇔(x-2)(x²+2x+4)-(x-2)(x+4)(x+10)=0

   ⇔(x-2)[x²+2x+4-(x+4)(x+10)]=0

   ⇔(x-2)(x²+2x+4-x²-10x-4x-40)=0

   ⇔(x-2)(-12x-36)=0

   ⇔-12(x-2)(x+3)=0

   ⇔\(\left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

   Vậy x=2;x=-3

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