Toán Lớp 8: D=((x^2-3x)/(x^2-9)-1):((9-x^2)/(x^2+x-6)-(x-3)/(2-x)-(x-2)/(x+3))
Rút gọn D rồi tìm x∈Z để D∈Z
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: D=((x^2-3x)/(x^2-9)-1):((9-x^2)/(x^2+x-6)-(x-3)/(2-x)-(x-2)/(x+3))
Rút gọn D rồi tìm x∈Z để D∈Z
Comments ( 1 )
x = 6\\
x = 0\\
x = 4
\end{array} \right.\)
D = \left( {\dfrac{{{x^2} – 3x}}{{{x^2} – 9}} – 1} \right):\left( {\dfrac{{9 – {x^2}}}{{{x^2} + x – 6}} – \dfrac{{x – 3}}{{2 – x}} – \dfrac{{x – 2}}{{x + 3}}} \right)\\
= \dfrac{{{x^2} – 3x – {x^2} + 9}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}:\dfrac{{9 – {x^2} + \left( {x – 3} \right)\left( {x + 3} \right) – {{\left( {x – 2} \right)}^2}}}{{\left( {x + 3} \right)\left( {x – 2} \right)}}\\
= \dfrac{{ – 3x + 9}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 3} \right)\left( {x – 2} \right)}}{{9 – {x^2} + {x^2} – 9 – {{\left( {x – 2} \right)}^2}}}\\
= – \dfrac{{3\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 3} \right)\left( {x – 2} \right)}}{{ – {{\left( {x – 2} \right)}^2}}}\\
= \dfrac{{3\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 3} \right)\left( {x – 2} \right)}}{{{{\left( {x – 2} \right)}^2}}}\\
= \dfrac{3}{{x – 2}}\\
D \in Z \to \dfrac{3}{{x – 2}} \in Z\\
\to x – 2 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x – 3 = 3\\
x – 3 = – 3\\
x – 3 = 1\\
x – 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = 0\\
x = 4\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)