Câu 1 :tìm x a,(2x-8)^2-6=0 b,(2x-1)^2-(4x+1)(z-3)=3 Câu 2: Cho x+y+z=0. Chứng minh rằng x^3+y^3+z^3=3xyz

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1. 1)
   \(\begin{array}{l}
   a){\left( {2x – 8} \right)^2} – 6 = 0\\
   \Leftrightarrow {\left( {2x – 8} \right)^2} = 6\\
   \Leftrightarrow \left[ \begin{array}{l}
   2x – 8 = \sqrt 6 \\
   2x – 8 = – \sqrt 6
   \end{array} \right.\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = \frac{{8 + \sqrt 6 }}{2}\\
   x = \frac{{8 – \sqrt 6 }}{2}
   \end{array} \right.\\
   b)\, \Leftrightarrow 4{x^2} – 4x + 1 – \left( {4{x^2} – 11x – 3} \right) = 3\\
   \Leftrightarrow 7x = 5 \Leftrightarrow x = \frac{5}{7}
   \end{array}\)
   2)
   Ta có \({\left( {x + y} \right)^3} = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} = {x^3} + {y^3} + 3xy\left( {x + y} \right)\)\( \Rightarrow {x^3} + {y^3} = {\left( {x + y} \right)^3} – 3xy\left( {x + y} \right)\)
   Xét \(y = {x^3} + {y^3} + {z^3} – 3xyz\)\( = {\left( {x + y} \right)^3} – 3xy\left( {x + y} \right) + {z^3} – 3xyz\)\( = \left[ {{{\left( {x + y} \right)}^3} + {z^3}} \right] – 3xy\left( {x + y + z} \right)\)
   \( = \left( {x + y + z} \right)\left[ {{{\left( {x + y} \right)}^2} – \left( {x + y} \right)z + {z^2}} \right] – 3xy\left( {x + y + z} \right)\)
   Mà \(x + y + z = 0\) nên \(y = 0.\left[ {{{\left( {x + y} \right)}^2} – \left( {x + y} \right)z + {z^2}} \right] – 3xy.0 = 0\)
   Vậy \(y = 0\) .
   Vậy \(x^3+y^3+z^3=3xyz\)

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