Giúp mik vs !
Cho a+b+c=0 (a,b,c khác 0)
Tính gtbt:
p= $\frac{a^2}{a^2-b^2-c^2}$ + $\frac{b^2}{b^2-c^2-a^2}$ + $\frac{c^2}{c^2 -a^2-b^2}$
Giúp mik vs !
Cho a+b+c=0 (a,b,c khác 0)
Tính gtbt:
p= $\frac{a^2}{a^2-b^2-c^2}$ + $\frac{b^2}{b^2-c^2-a^2}$ + $\frac{c^2}{c^2 -a^2-b^2}$
$\begin{array}{l}\text{Đáp án:}\\P=\dfrac{3}{2}\\\text{Giải thích các bước giải:}\\a+b+c=0\\\to \begin{cases}a+b=-c\\b+c=-a\\c+a=-b\\\end{cases}\\P=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\\=\dfrac{a^2}{(a-b)(a+b)-c^2}+\dfrac{b^2}{(b-c)(c+b)-a^2}+\dfrac{c^2}{(c-a)(c+a)-b^2}\\=\dfrac{a^2}{-c(a-b)-c^2}+\dfrac{b^2}{-a(b-c)-a^2}+\dfrac{c^2}{-b(c-a)-b^2}\\=\dfrac{a^2}{-c(a-b+c)}+\dfrac{b^2}{-a(b-c+a)}+\dfrac{c^2}{-b(c-a+b)}\\=\dfrac{a^2}{-c(-2b)}+\dfrac{b^2}{-a(-2c)}+\dfrac{c^2}{-b(-2a)}\\=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\\=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\\=\dfrac{a^3+b^3+c^3}{2abc}(1)\\\text{Mặt khác:}\\a+b+c=0\\\to a+b=-c\\\to (a+b)^3=(-c)^3\\\to a^3+3a^2b+3ab^2+b^3=-c^3\\\to a^3+b^3+c^3=-3a^b-3ab^2\\\to a^3+b^3+c^3=-3ab(a+b)\\\to a^3+b^3+c^3=-3ab(-c)\\\to a^3+b^3+c^3=3abc(2)\\\text{Từ (1) và (2) suy ra:}P=\dfrac{3abc}{2abc}=\dfrac{3}{2}\\\end{array}$