giải pt: a) √(x ² – 1) + √(x – 1) = 0 b) √( x ² – 9) + √( x- 3) = 0 c) √(x ² – 3x + 2) + √(x – 1) =0 d) √(x ² – 5x + 4) + √(x – 4) = 0

0 bình luận về “giải pt: a) √(x ² – 1) + √(x – 1) = 0 b) √( x ² – 9) + √( x- 3) = 0 c) √(x ² – 3x + 2) + √(x – 1) =0 d) √(x ² – 5x + 4) + √(x – 4) = 0”

1. Đáp án:

   a, Ta có

   $\sqrt{x^2 – 1 }  + \sqrt{x – 1 }  = 0$ `(ĐKXĐ : x ≥ 1)`

   ` <=> x^2 – 1 + (x – 1) = 0`

   ` <=> (x – 1)(x + 1) + (x – 1) = 0`

   ` <=> (x – 1)(x + 2) = 0`

   <=> \(\left[ \begin{array}{l}x – 1 = 0\\x + 2 = 0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=1\\x=-2 < Loại >\end{array} \right.\) 

   ` <=> x = 1`

   b, Ta có

   $\sqrt{x^2 – 9}$  + $\sqrt{x – 3 }$  = 0  `(ĐKXĐ : x ≥ 3)`

   ` <=> x^2 – 9 + (x – 3) = 0`

   ` <=> (x – 3)(x + 3) + (x – 3) = 0`

   ` <=> (x – 3)(x + 4) = 0`

   <=> \(\left[ \begin{array}{l}x – 3 = 0\\x + 4 = 0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=3\\x=-4 < Loại >\end{array} \right.\) 

   ` <=> x = 3`

   c, Ta có

   $\sqrt{x^2 – 3x + 2}$  + $\sqrt{x – 1 }$  = 0 `(ĐKXĐ : x ≥ 1 )`

   ` <=> x^2 – 3x + 2 + x – 1 = 0`

   ` <=> x^2 – x – 2x + 2 + (x – 1) = 0`

   ` <=> x(x – 1) – 2(x – 1) + (x – 1) = 0`

   ` <=> (x – 1)(x – 1) = 0`

   ` <=> x – 1 = 0`

   ` <=> x = 1`

   d, Ta có

   $\sqrt{x^2 – 5x + 4}$  + $\sqrt{x – 4 }$  = 0  `(ĐKXĐ : x ≥ 4)`

   ` <=> x^2 – 5x + 4 + x – 4 = 0`

   ` <=> x^2 – x – 4x + 4 + (x – 4) = 0`

   ` <=> x(x – 1) – 4(x – 1) + (x – 4) = 0`

   ` <=> (x – 1)(x – 4) + (x – 4) = 0`

   ` <=> (x – 4)x = 0`

   <=> \(\left[ \begin{array}{l}x – 4 = 0\\x=0\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x=4\\x=0\end{array} \right.\) 

   ` <=> x = 4`

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