(2x+3)^2-25=0 9x^2-(x+2)=0 (2x-1)^2-(x-4)^2=0 (2x-1)^2-(x-4)=0 X^3+x^2-4x-4=0

0 bình luận về “(2x+3)^2-25=0 9x^2-(x+2)=0 (2x-1)^2-(x-4)^2=0 (2x-1)^2-(x-4)=0 X^3+x^2-4x-4=0”

1. Giải thích các bước giải:

   $(2x +3)² -25 = 0$

   `⇔ (2x+3)^2 – 5^2=0`
   $⇔ (2x +3 -5).(2x +3 +5) = 0$
   $⇔ (2x -2).(2x +8) = 0$
   $⇔ \left[ \begin{array}{l}2x -2=0\\2x +8=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=-4\end{array} \right.$

   Vậy `x ∈ {1;-4}`

   _________

   $9x² -(x +2)² = 0$
   $⇔ (3x -x -2).(3x +x +2) = 0$
   $⇔ (2x -2).(4x +2) = 0$
   $⇔ \left[ \begin{array}{l}2x -2=0\\4x +2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=\dfrac{-1}{2}\end{array} \right.$

   Vậy `x ∈ {1; -1/2}`

   _________

   $(2x -1)² -(x -4)² = 0$
   $⇔ (2x -1 -x +4).(2x -1 +x -4) = 0$
   $⇔ (x +3).(3x -5) = 0$
   $⇔ \left[ \begin{array}{l}x +3=0\\3x -5=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-3\\x=\dfrac{5}{3}\end{array} \right.$

   Vậy `x ∈ {-3; 5/3}`

   _________

   $(2x -1)² -(x -3)² = 0$
   $⇔ (2x -1 -x +3).(2x -1 +x -3) = 0$
   $⇔ (x +2).(3x -4) = 0$
   $⇔ \left[ \begin{array}{l}x +2=0\\3x -4=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-2\\x=\dfrac{4}{3}\end{array} \right.$

   Vậy `x ∈ {-2; 4/3}`

   __________

   $x³ +x² -4x -4 = 0$
   `⇔x^2(x+1)-4(x+1)=0`

   `⇔ (x^2-4)(x+1)=0`

   \(\Leftrightarrow\left[ \begin{array}{l}x^2-4=0\\x+1=0\end{array} \right.\) \(\Leftrightarrow\left[ \begin{array}{l}x^2=4\\x=-1\end{array} \right.\) \(\Leftrightarrow\left[ \begin{array}{l}x=\pm2\\x=-1\end{array} \right.\)

   Vậy `x ∈ {±2;-1}`

   ____________

   `(2x + 1)^2 – (x – 4)^2=0`

   `⇔ (2x + 1+ x – 4)(2x+1-x+4)=0`

   `⇔ (3x-3)(x+5)=0`

   \(\Leftrightarrow\left[ \begin{array}{l}3x-3=0\\x+5=0\end{array} \right.\) \(\Leftrightarrow\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\)

   Vậy `x ∈ {1;-5}`

   __________

   `x^3 – x^2 – x + 1=0`

   `⇔ x^2(x – 1)-(x-1)=0`

   `⇔ (x^2 – 1)(x-1)=0`

   \(\Leftrightarrow\left[ \begin{array}{l}x^2-1=0\\x-1=0\end{array} \right.\) \(\Leftrightarrow\left[ \begin{array}{l}x^2=1\\x=1\end{array} \right.\) \(\Leftrightarrow\left[ \begin{array}{l}x=\pm1\\x=1\end{array} \right.\)

   Vậy `x =±1`

   __________

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2. Giải thích các bước giải:

    $a) (2x +3)² -25 = 0$

   $⇔ (2x +3 -5).(2x +3 +5) = 0$

   $⇔ (2x -2).(2x +8) = 0$

   $⇔ \left[ \begin{array}{l}2x -2=0\\2x +8=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=-4\end{array} \right.$

   Vậy `S = {1; -4}`

   $b) 9x² -(x +2)² = 0$

   $⇔ (3x -x -2).(3x +x +2) = 0$

   $⇔ (2x -2).(4x +2) = 0$

   $⇔ \left[ \begin{array}{l}2x -2=0\\4x +2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=\dfrac{-1}{2}\end{array} \right.$

   Vậy `S = {1; -1/2}`

   $c) (2x -1)² -(x -4)² = 0$

   $⇔ (2x -1 -x +4).(2x -1 +x -4) = 0$

   $⇔ (x +3).(3x -5) = 0$

   $⇔ \left[ \begin{array}{l}x +3=0\\3x -5=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-3\\x=\dfrac{5}{3}\end{array} \right.$

   Vậy `S = {-3; 5/3}`

   $d) (2x -1)² -(x -3)² = 0$

   $⇔ (2x -1 -x +3).(2x -1 +x -3) = 0$

   $⇔ (x +2).(3x -4) = 0$

   $⇔ \left[ \begin{array}{l}x +2=0\\3x -4=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-2\\x=\dfrac{4}{3}\end{array} \right.$

   Vậy `S = {-2; 4/3}`

   $e) x³ +x² -4x -4 = 0$

   $⇔ x.(x +1) -4.(x +1) = 0$

   $⇔ (x +1).(x -4) = 0$

   $⇔ \left[ \begin{array}{l}x +1=0\\x -4=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-1\\x=4\end{array} \right.$

   Vậy `S = {-1; 4}`

   $g) (2x+1)² -(x-4)² =0 $

   $⇔ (2x +1 -x +4).(2x +1 +x -4) = 0$

   $⇔ (x +5).(3x -3) = 0$

   $⇔ \left[ \begin{array}{l}x +5=0\\3x -3=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-5\\x=1\end{array} \right.$

   Vậy `S = {-5; 1}`

   $l) x³ -x² -x +1 = 0$

   $⇔ x².(x -1) -(x -1) = 0$

   $⇔ (x -1).(x² -1) = 0$

   $⇔ \left[ \begin{array}{l}x -1=0\\x² -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=±1\end{array} \right.$

   Vậy `S = {±1}`

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