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http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/
Ta có: \(a+b+c=0\)
\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)
\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)
\(=\frac{a^3+b^3+c^3}{2abc}=\frac{a^3+b^3+c^3-3abc+3abc}{2abc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)
\(=\frac{3}{2}\)
ĐK : a;b;c khác 0
Thấy : \(a^2+b^2+c^2=\left(a+b+c\right)^2\Leftrightarrow ab+bc+ac=0\) (1)
Ta có : \(P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)
Từ (1) suy ra : \(\left(b+c\right)a=-bc\Leftrightarrow\dfrac{b+c}{a}=\dfrac{-bc}{a^2}\)
CMTT ; ta có : \(\dfrac{c+a}{b}=\dfrac{-ac}{b^2};\dfrac{a+b}{c}=\dfrac{-ab}{c^2}\)
Suy ra : \(P=-\left(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}\right)=-\dfrac{a^3b^3+b^3c^3+a^3c^3}{a^2b^2c^2}\) (2)
Đặt : ab = x ; bc = y ; ac = z ; ta có : x + y + z = 0 \(\Rightarrow x^3+y^3+z^3=3xyz\) (3)
Từ (2) và (3) suy ra : \(P=-\dfrac{3xyz}{xyz}=-3\)
Vậy ...
P= \(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0
Ngu vãi ko bt đúng không nx
Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow ab+bc+ac=0$
Đặt $ab=x, bc=y, ac=z$ thì $x+y+z=0$
Có:
$M=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}$
$=\frac{b^3c^3+a^3c^3+a^3b^3}{(abc)^2}$
$=\frac{x^3+y^3+z^3}{xyz}=\frac{(x+y)^3-3xy(x+y)+z^3}{xyz}$
$=\frac{(-z)^3-3xy(-z)+z^3}{xyz}$
$+\frac{-z^3+3xyz+z^3}{xyz}=\frac{3xyz}{xyz}=3$
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Mà \(a+b+c\ne0\left(gt\right)\)
\(\Leftrightarrow a=b=c\)
Do đó:
\(A=\frac{a^2+2b^2+6c^2}{\left(a+b+c\right)^2}+2015=\frac{a^2+2a^2+6c^2}{\left(a+a+a\right)^2}+2015=\frac{9a^2}{9a^2}+2015=1+2015=2016\)
a+b+c=0
=>a+b=-c; a+c=-b; b+c=-a
\(a^2-b^2-c^2\)
\(=a^2-\left(b^2+c^2\right)\)
\(=a^2-\left\lbrack\left(b+c\right)^2-2bc\right\rbrack=a^2-\left(a^2-2bc\right)=2bc\)
\(b^2-a^2-c^2\)
\(=b^2-\left(a^2+c^2\right)\)
\(=b^2-\left\lbrack\left(a+c\right)^2-2ac\right\rbrack=b^2-\left(b^2-2ac\right)=2ac\)
\(c^2-a^2-b^2\)
\(=c^2-\left\lbrack a^2+b^2\right\rbrack\)
\(=c^2-\left\lbrack\left(a+b\right)^2-2ab\right\rbrack=c^2-\left(c^2-2ab\right)=2ab\)
Ta có: \(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
\(=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2bac}=\frac{\left(-c\right)^3-3ab\cdot\left(-c\right)+c^3}{2abc}=\frac{3abc}{2abc}=\frac32\)