Tìm x 1, (x-1)(x^2 + x +1) – x(x+2)(x-2) = 5 2, (x+1)^2 – 2(x+1)(3x -2) + (3x -2)^2 =0 3, x^2 – 2x +1=0 4, (5x +1)^2 – (5x -3)(5x+3)=30

0 bình luận về “Tìm x 1, (x-1)(x^2 + x +1) – x(x+2)(x-2) = 5 2, (x+1)^2 – 2(x+1)(3x -2) + (3x -2)^2 =0 3, x^2 – 2x +1=0 4, (5x +1)^2 – (5x -3)(5x+3)=30”

1. Đáp án:

   Giải thích các bước giải:

   \(\begin{array}{l}
   1.\,\,\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) – x\left( {x + 2} \right)\left( {x – 2} \right) = 5\\
   \Leftrightarrow {x^3} – 1 – x\left( {{x^2} – 4} \right) = 5\\
   \Leftrightarrow 4x – 1 = 5\\
   \Leftrightarrow 4x = 6\\
   \Leftrightarrow x = \frac{3}{2}\\
   2.\,\,\,{\left( {x + 1} \right)^2} – 2\left( {x + 1} \right)\left( {3x – 2} \right) + {\left( {3x – 2} \right)^2} = 0\\
   \Leftrightarrow {\left( {x + 1 – 3x + 2} \right)^2} = 0\\
   \Leftrightarrow – 2x + 3 = 0\\
   \Leftrightarrow x = \frac{3}{2}\\
   3.\,\,\,{x^2} – 2x + 1 = 0\\
   \Leftrightarrow {\left( {x – 1} \right)^2} = 0\\
   \Leftrightarrow x – 1 = 0\\
   \Leftrightarrow x = 1\\
   4.\,\,{\left( {5x + 1} \right)^2} – \left( {5x – 3} \right)\left( {5x + 3} \right) = 30\\
   \Leftrightarrow 25{x^2} + 10x + 1 – 25{x^2} + 9 = 30\\
   \Leftrightarrow 10x + 10 = 30\\
   \Leftrightarrow 10x = 20\\
   \Leftrightarrow x = 2
   \end{array}\)

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