Bài 1 Phân tích
1.(a+b+c)²+(a+b-c)²-4c²
2.4a²b²-(a²+b²-c²)²
3.a(b²-c²)+b(c²-a²)+c(a²-b²)
4.ab(a+b)+bc(b+c)+ca(c+a)+2abc
5.ab(a+b)+bc(b+c)+ca(c+a)+3abc
Bài 1 Phân tích
1.(a+b+c)²+(a+b-c)²-4c²
2.4a²b²-(a²+b²-c²)²
3.a(b²-c²)+b(c²-a²)+c(a²-b²)
4.ab(a+b)+bc(b+c)+ca(c+a)+2abc
5.ab(a+b)+bc(b+c)+ca(c+a)+3abc
Giải thích các bước giải:
$\begin{array}{l}
1){\left( {a + b + c} \right)^2} + {\left( {a + b – c} \right)^2} – 4{c^2}\\
= {\left( {a + b + c} \right)^2} – {\left( {2c} \right)^2} + {\left( {a + b – c} \right)^2}\\
= \left( {a + b – c} \right)\left( {a + b + 3c} \right) + {\left( {a + b – c} \right)^2}\\
= \left( {a + b – c} \right)\left( {a + b + 3c + a + b – c} \right)\\
= \left( {a + b – c} \right)\left( {2a + 2b + 2c} \right)\\
= 2\left( {a + b – c} \right)\left( {a + b + c} \right)\\
2)4{a^2}{b^2} – {\left( {{a^2} + {b^2} – {c^2}} \right)^2}\\
= {\left( {2ab} \right)^2} – {\left( {{a^2} + {b^2} – {c^2}} \right)^2}\\
= \left( {2ab – \left( {{a^2} + {b^2} – {c^2}} \right)} \right)\left( {2ab + {a^2} + {b^2} – {c^2}} \right)\\
= \left( {{c^2} – \left( {{a^2} + {b^2} – 2ab} \right)} \right)\left( {\left( {{a^2} + {b^2} + 2ab} \right) – {c^2}} \right)\\
= \left( {{c^2} – {{\left( {a – b} \right)}^2}} \right)\left( {{{\left( {a + b} \right)}^2} – {c^2}} \right)\\
= \left( {c – a + b} \right)\left( {c + a – b} \right)\left( {a + b – c} \right)\left( {a + b + c} \right)\\
= \left( { – a + b + c} \right)\left( {a – b + c} \right)\left( {a + b – c} \right)\left( {a + b + c} \right)\\
3)a\left( {{b^2} – {c^2}} \right) + b\left( {{c^2} – {a^2}} \right) + c\left( {{a^2} – {b^2}} \right)\\
= a\left( {b – c} \right)\left( {b + c} \right) + b{c^2} – {a^2}b + {a^2}c – {b^2}c\\
= \left( {b – c} \right)\left( {ab + ac} \right) – bc\left( {b – c} \right) – {a^2}\left( {b – c} \right)\\
= \left( {b – c} \right)\left( {ab + ac – bc – {a^2}} \right)\\
= \left( {b – c} \right)\left( {b\left( {a – c} \right) – a\left( {a – c} \right)} \right)\\
= \left( {b – c} \right)\left( {a – c} \right)\left( {b – a} \right)\\
4)ab\left( {a + b} \right) + bc\left( {b + c} \right) + ca\left( {c + a} \right) + 2abc\\
= ab\left( {a + b} \right) + abc + bc\left( {b + c} \right) + abc + ca\left( {c + a} \right)\\
= ab\left( {a + b + c} \right) + bc\left( {a + b + c} \right) + ca\left( {c + a} \right)\\
= \left( {a + b + c} \right)\left( {ab + bc} \right) + ca\left( {c + a} \right)\\
= \left( {a + b + c} \right)b\left( {a + c} \right) + ca\left( {c + a} \right)\\
= \left( {a + c} \right)\left( {b\left( {a + b + c} \right) + ca} \right)\\
= \left( {a + c} \right)\left( {ab + {b^2} + bc + ca} \right)\\
= \left( {a + c} \right)\left( {a + b} \right)\left( {b + c} \right)\\
5)ab\left( {a + b} \right) + bc\left( {b + c} \right) + ca\left( {c + a} \right) + 3abc\\
= \left( {ab\left( {a + b} \right) + abc} \right) + \left( {bc\left( {b + c} \right) + abc} \right) + \left( {ca\left( {c + a} \right) + abc} \right)\\
= \left( {a + b + c} \right)\left( {ab + bc + ca} \right)
\end{array}$