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+

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)

\(a+b+c=0\)

\(a+b=-c\)

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

\(a^3+3a^2b+3ab^2+b^3=-c^3\)

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

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

\(a^3+b^3+c^3=3abc\left(1\right)\)

\(P=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}\)

\(P=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(P=\dfrac{a^3+b^3+c^3}{abc}\)

Thay (1) vào P ta được :

\(P=\dfrac{3abc}{abc}=3\)

Vậy.......

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

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

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

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

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

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có VT:

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)

\(=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\) (1)

VT: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\) (2)

Từ (1) và (2) 

\(\Rightarrow\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\left(đpcm\right)\)

Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ab=cd\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)\(\Leftrightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)

Vậy...

P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )

Ta có : 

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

Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)

\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)

Ta có: \(a+b+c=0\Rightarrow a^2=\left(b+c\right)^2\Rightarrow a^2-2bc=b^2+c^2\)

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

Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

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

Lại có: \(a+b+c=0\Rightarrow-a=b+c\)

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

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

=> \(A=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)