a) (x+1)^3+(x-2)^3-2x^2(x-1,5)=3 b) (x+4)(x^2-4x+16)-x(x-5)(x+5)=264 c) (x-2)^3-(x-2)(x^2+2x+4)+6(x-2)(x+2)=60

0 bình luận về “a) (x+1)^3+(x-2)^3-2x^2(x-1,5)=3 b) (x+4)(x^2-4x+16)-x(x-5)(x+5)=264 c) (x-2)^3-(x-2)(x^2+2x+4)+6(x-2)(x+2)=60”

1. Đáp án:

   $\begin{array}{l}
   a){\left( {x + 1} \right)^3} + {\left( {x – 2} \right)^3} – 2{x^2}\left( {x – 1,5} \right) = 3\\
    \Rightarrow {x^3} + 3{x^2} + 3x + 1 + \\
   {x^3} – 6{x^2} + 12x – 8 – 2{x^3} + 3{x^2} = 3\\
    \Rightarrow 15x = 10\\
    \Rightarrow x = \frac{2}{3}\\
   b)\left( {x + 4} \right)\left( {{x^2} – 4x + 16} \right) – x\left( {x – 5} \right)\left( {x + 5} \right) = 264\\
    \Rightarrow {x^3} + {4^3} – x\left( {{x^2} – 25} \right) = 264\\
    \Rightarrow {x^3} + 64 – {x^3} + 25x – 264 = 0\\
    \Rightarrow 25x – 200 = 0\\
    \Rightarrow x = 8\\
   c){\left( {x – 2} \right)^3} – \left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right) + \\
   6\left( {x – 2} \right)\left( {x + 2} \right) = 60\\
    \Rightarrow {x^3} – 6{x^2} + 12x – 8 – \left( {{x^3} – {2^3}} \right)\\
    + 6\left( {{x^2} – 4} \right) = 60\\
    \Rightarrow {x^3} – 6{x^2} + 12x – 8 – {x^3} + 8 + 6{x^2} – 24 = 60\\
    \Rightarrow 12x = 84\\
    \Rightarrow x = 7
   \end{array}$

   Vậy x=7

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