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

0 bình luận về “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”

  1. 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}$

    Bình luận

Viết một bình luận