Toán Lớp 9: giúp em với ạ Rút gọn P= a^2(b-c)+b^2(c-a)+c^2(a-b)/ab^2-ac^2-b^3+bc^2
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 9: giúp em với ạ Rút gọn P= a^2(b-c)+b^2(c-a)+c^2(a-b)/ab^2-ac^2-b^3+bc^2
Comments ( 1 )
P = \dfrac{{{a^2}\left( {b – c} \right) + {b^2}\left( {c – a} \right) + {c^2}\left( {a – b} \right)}}{{a{b^2} – a{c^2} – {b^3} + b{c^2}}}\\
= \dfrac{{{a^2}b – {a^2}c + {b^2}c – a{b^2} + {c^2}\left( {a – b} \right)}}{{a\left( {{b^2} – {c^2}} \right) – b\left( {{b^2} – {c^2}} \right)}}\\
= \dfrac{{{a^2}b – a{b^2} – \left( {{a^2}c – {b^2}c} \right) + {c^2}\left( {a – b} \right)}}{{\left( {a – b} \right)\left( {{b^2} – {c^2}} \right)}}\\
= \dfrac{{ab\left( {a – b} \right) – c\left( {{a^2} – {b^2}} \right) + {c^2}\left( {a – b} \right)}}{{\left( {a – b} \right)\left( {b – c} \right)\left( {b + c} \right)}}\\
= \dfrac{{\left( {a – b} \right)\left( {ab – c\left( {a + b} \right) + {c^2}} \right)}}{{\left( {a – b} \right)\left( {b – c} \right)\left( {b + c} \right)}}\\
= \dfrac{{ab – ac – bc + {c^2}}}{{\left( {b – c} \right)\left( {b + c} \right)}}\\
= \dfrac{{b\left( {a – c} \right) – c\left( {a – c} \right)}}{{\left( {b – c} \right)\left( {b + c} \right)}}\\
= \dfrac{{\left( {a – c} \right)\left( {b – c} \right)}}{{\left( {b – c} \right)\left( {b + c} \right)}}\\
= \dfrac{{a – c}}{{b + c}}
\end{array}$