Thực hiện phép tính: a) $\frac{2x+9}{x-5}$ . $\frac{5x-8}{x+1945}$ – $\frac{2x+9}{x-5}$ . $\frac{4x-3}{x+1945}$ b) $\frac{x^3+y^3+z^3-3xyz}{xyt^2+x

Đáp án:

 a) \(\dfrac{{2x + 9}}{{x + 1945}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { – 1945;5} \right\}\\
\dfrac{{\left( {2x + 9} \right)\left( {5x – 8} \right) – \left( {2x + 9} \right)\left( {4x – 3} \right)}}{{\left( {x – 5} \right)\left( {x + 1945} \right)}}\\
 = \dfrac{{\left( {2x + 9} \right)\left( {5x – 8 – 4x + 3} \right)}}{{\left( {x – 5} \right)\left( {x + 1945} \right)}}\\
 = \dfrac{{\left( {2x + 9} \right)\left( {x – 5} \right)}}{{\left( {x – 5} \right)\left( {x + 1945} \right)}}\\
 = \dfrac{{2x + 9}}{{x + 1945}}\\
b){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)^3} – 3z\left( {x + y} \right)\left( {x + y + z} \right) – 3xy\left( {x – y – z} \right)\\
\; = \left( {x + y + z} \right)\left[ {{{\left( {x + y + z} \right)}^2} – 3z\left( {x + y} \right) – 3xy} \right]\;\\
 = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} + 2xy + 2xz + 2yz – 3xz – 3yz – 3xy} \right)\\
 = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} – xy – xz – yz} \right)\\
{\left( {x – y} \right)^2} + {\left( {y – z} \right)^2} + {\left( {x – z} \right)^2}\\
 = {x^2} – 2xy + {y^2} + {y^2} – 2yz + {z^2} + {x^2} – 2xz + {z^2}\\
 = 2{x^2} + 2{y^2} + 2{z^2} – 2xy – 2yz – 2xz\\
 = 2\left( {{x^2} + {y^2} + {z^2} – xy – xz – yz} \right)\\
\dfrac{{{x^3} + {y^3} + {z^3} – 3xyz}}{{xy{z^2} + xz\left( {zy + z} \right)}}.\dfrac{{x\left( {{y^2} + z} \right) + y\left( {x – xy} \right)}}{{{{\left( {x – y} \right)}^2} + {{\left( {y – z} \right)}^2} + {{\left( {x – z} \right)}^2}}}\\
 = \dfrac{{\left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} – xy – xz – yz} \right)}}{{xz\left( {yz + yz + z} \right)}}.\dfrac{{x{y^2} + xz + xy – x{y^2}}}{{2\left( {{x^2} + {y^2} + {z^2} – xy – xz – yz} \right)}}\\
 = \dfrac{{x + y + z}}{{x{z^2}\left( {2y + 1} \right)}}.\dfrac{{x\left( {y + z} \right)}}{2}\\
 = \dfrac{{\left( {x + y + z} \right)\left( {y + z} \right)}}{{2{z^2}\left( {2y + 1} \right)}}
\end{array}\)