Ta có a+b+c=0 => b+c=-a => a^2=b^2+2bc+c^2=> a^2-b^2-c^2=2bc

Tương tự ta có : b^2-c^2-a^2=2ca

c^2-a^2-b^2=2ab

=> a^2/2bc+b^2/2ca+c^2/2ab=(a^3+b^3+c^3)/2abc

=>Ta lại có a^3+b^3+c^3=(a+b+c)^3+

Từ giả thiết ta có:

\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^2=a^2\)

\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)

Tương tự:

\(b^2-c^2-a^2=2ca,c^2-a^2-b^2=2ab\)

Từ đây suy ra:

\(A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{2abc}\)

Mặt khác lại có:

\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^3=-a^3\)

\(\Rightarrow b^3+c^3+3bc\left(b+c\right)=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow A=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3}{2}\)

Ngô Tấn Đạt

Ngô Thanh Sang

Ta có: a+b+c=0

<=>a=-b-c

<=>\(a^2=(-b-c)^2\)

<=>\(a^2=b^2+c^2+2bc\)

cmtt:\(b^2=a^2+c^2+2ac\)

\(c^2=a^2+b^2+2ab\)

=> A=\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3}{2abc}+\frac{b^3}{ 2abc}+\frac{c^3}{2abc} \)

=\(\frac{1}{2abc}(a^3+b^3+c^ 3)\)

Cm đẳng thức phụ

Với a+b+c=0=> \(a^3+b^3+c^3=3abc\)

=>A=\(\frac{3}{2} \)

Áp dụng t/chất dãy tỉ số bằng nhau :

\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{\left(a+b+c\right)-\left(a-b+c\right)}{\left(a+b-c\right)-\left(a-b-c\right)}=\frac{2b}{2b}=1\) (bỏ dấu ngoặc)

\(\Rightarrow a+b+c=a+b-c\Rightarrow c=-c\Rightarrow2c=0\Rightarrow c=0\) (đpcm)

Bài 11 là \(a+b+c=0\)thôi nha, không có a;b;c khác 0 đâu tui bị nhầm đó, xin lỗi nhiều ;;;

j giàu thế :) cho xin ít đi

\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)

Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)

Từ đó tự giải ra

Áp dụng t/c dtsbn:

\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{b+c-5+a+c+2+a+b+3}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\)\(\left(1\right)\)

Mặt khác \(\dfrac{1}{a+b+c}=\dfrac{b+c-5}{a}=2\)\(\Rightarrow a+b+c=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{2}-c\\a+c=\dfrac{1}{2}-b\\b+c=\dfrac{1}{2}-a\end{matrix}\right.\)\(\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-a-5=2a\\\dfrac{1}{2}-b+2=2b\\\dfrac{1}{2}-c+3=2c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{2}\\b=\dfrac{5}{6}\\c=\dfrac{7}{6}\end{matrix}\right.\)

\(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)=\left(-\dfrac{3}{2}-3.\dfrac{5}{6}\right)\left(\dfrac{5}{6}-\dfrac{7}{6}\right)\left(3.\dfrac{7}{6}+\dfrac{3}{2}\right)=\dfrac{20}{3}\)

Lời giải:

Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$

$\Rightarrow x=at; y=bt; z=ct$. Ta có:

$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$

Mặt khác:

$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$

Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)