# Toán Lớp 9: A=(a^3)/((a-b)(a-c))+(b^3)/((b-a)(b-c))+(c^3)/((c-a)(c-b))

Toán Lớp 9: A=(a^3)/((a-b)(a-c))+(b^3)/((b-a)(b-c))+(c^3)/((c-a)(c-b))

### 0 bình luận về “Toán Lớp 9: A=(a^3)/((a-b)(a-c))+(b^3)/((b-a)(b-c))+(c^3)/((c-a)(c-b))”

1. Giải đáp:

Lời giải và giải thích chi tiết:
A=(a^3)/((a-b)(a-c))+(b^3)/((b-a)(b-c))+(c^3)/((c-a)(c-b))
ĐK:a \ne b \ne c
A=(a^3)/((a-b)(a-c))-(b^3)/((a-b)(b-c))+(c^3)/((a-c)(b-c))
A=(a^3(b-c))/((a-b)(a-c)(b-c))-(b^3(a-c))/((a-b)(a-c)(b-c))+(c^3(a-b))/((a-b)(a-c)(b-c))
A=(a^3(b-c)-b^3(a-c)+c^3(a-b))/((a-b)(a-c)(b-c))
A=((a-b)(a-c)(b-c)(a+b+c))/((a-b)(a-c)(b-c))
A=(a+b+c)

Trả lời
2. $\begin{array}{l} \dfrac{{{a^3}}}{{\left( {a – b} \right)\left( {a – c} \right)}} + \dfrac{{{b^3}}}{{\left( {b – c} \right)\left( {b – a} \right)}} + \dfrac{{{c^3}}}{{\left( {c – b} \right)\left( {c – a} \right)}}\\ = \dfrac{{{a^3}\left( {c – b} \right) + {b^3}\left( {a – c} \right) + {c^3}\left( {b – a} \right)}}{{\left( {a – b} \right)\left( {b – c} \right)\left( {c – a} \right)}}\\ = \dfrac{{{a^3}c – {c^3}a + {b^3}a – {a^3}b + {c^3}b – {b^3}c}}{{\left( {a – b} \right)\left( {b – c} \right)\left( {c – a} \right)}}\\ TS:\\ {a^3}c – {c^3}a + {b^3}a – {a^3}b + {c^3}b – {b^3}c\\ = ac\left( {{a^2} – {c^2}} \right) + {b^3}\left( {a – c} \right) – b\left( {{a^3} – {c^3}} \right)\\ = ac\left( {a – c} \right)\left( {a + c} \right) + {b^3}\left( {a – c} \right) – b\left( {a – c} \right)\left( {{a^2} + ac + {c^2}} \right)\\ = \left( {a – c} \right)\left( {ac\left( {a + c} \right) + {b^3} – b\left( {{a^2} + ac + {c^2}} \right)} \right)\\ = \left( {a – c} \right)\left( {{a^2}c + a{c^2} + {b^3} – {a^2}b – abc – b{c^2}} \right)\\ = \left( {a – c} \right)\left[ {ac\left( {a – b} \right) + {c^2}\left( {a – b} \right) – b\left( {{a^2} – {b^2}} \right)} \right]\\ = \left( {a – c} \right)\left( {a – b} \right)\left( {ac + {c^2} – b\left( {a + b} \right)} \right)\\ = \left( {a – c} \right)\left( {a – b} \right)\left( {ac + {c^2} – ab – {b^2}} \right)\\ = \left( {a – c} \right)\left( {a – b} \right)\left[ {a\left( {c – b} \right) + \left( {c – b} \right)\left( {c + b} \right)} \right]\\ = \left( {a – c} \right)\left( {a – b} \right)\left( {c – b} \right)\left( {a + b + c} \right)\\ = \left( {c – a} \right)\left( {b – c} \right)\left( {a – b} \right)\left( {a + b + c} \right)\\ \Rightarrow A = a + b + c \end{array}$

Trả lời