Tìm x: a, $(x-5)^{2}$ = $(1-3x)^{2}$ b, $(3y-1)^{10}$ = $(3y-1)^{20}$

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1. a) `(x-5)^2 = (1-3x)^2`

   +) `x-5 = 1-3x`

   `=> x + 3x =1+5`

   `=> 4x= 6`

   `=> x= 3/2`

   +) `x -5 = 3x-1`

   `=> x- 3x =-1+5`

   `=> -2x = 4`

   `=> x= = 4:(-2)`

   `=> x= -2`

   Vậy `x= -2` hoặc `x= 3/2`

   b) `(3y-1)^10 = (3y-1)^20`

   `=> (3y-1)^10 – (3y-1)^20 =0`

   `=> (3y-1)^10 [1 – ( 3y-1)^2 ] =0`

   `=>` \(\left[ \begin{array}{l}3y-1 =0\\1-(3y-1)^2=0\end{array} \right.\) 

   `=>` \(\left[ \begin{array}{l}y=\frac{1}{3}\\(3y-1)^2=1\end{array} \right.\)

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

   `=>` \(\left[ \begin{array}{l}y=\frac{1}{3}\\y=\frac{2}{3}\\y=0\end{array} \right.\)

   Vậy `y = 1/3` hoặc `y= 2/3` hoặc `y=0`

    

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2. Đáp án:

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

   `+)x-5=1-3x`

   `<=>4x=6`

   `<=>x=3/2`.

   `+)x-5=3x-1`

   `<=>2x=-4`

   `<=>x=-2`.

   Vậy `x=3/2` hoặc `x=-2`.

   `b)(3y-1)^2=(3y-1)^20`

   `<=>(3y-1)^20-(3y-1)^2=0`

   `<=>(3y-1)^2[(3y-1)^18-1]=0`

   `+)(3y-1)^2=0`

   `<=>3y-1=0`

   `<=>3y=1`

   `<=>y=1/3`.

   `+)(3y-1)^18-1=0`

   `<=>(3y-1)^18=1`

   `++)3y-1=1`

   `<=>3y=2`

   `<=>y=2/3`.

   `++)3y-1=-1`

   `<=>3y=0`

   `<=>y=0`.

   Vậy `y=1/3` hoặc `y=2/3` hoặc `y=0`.

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