\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

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

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)    (do a+b+c = 0)

=>  \(B=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{ \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

=>   đpcm

1) \(a+b+c=0\Rightarrow2\left(a+b+c\right)=0\Rightarrow\frac{2\left(a+b+c\right)}{abc}=0\)

\(\Rightarrow M=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(x+y+z\right)}{xyz}\)

\(\Rightarrow M=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{yz}+\frac{2}{zx}+\frac{2}{xy}\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

3/ Ta có:

\(x+y+z=0\)

\(\Rightarrow x^2=\left(y+z\right)^2;y^2=\left(z+x\right)^2;z^2=\left(x+y\right)^2\)

\(a+b+c=0\)

\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

Ta có:

\(ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)

\(=x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\)

\(=-ax^2-by^2-cz^2\)

\(\Leftrightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Leftrightarrow ax^2+by^2+cz^2=0\)

1/ Đặt \(a-b=x,b-c=y,c-z=z\)

\(\Rightarrow x+y+z=0\)

Ta có:

\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)

\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

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

\(\Leftrightarrow ab+bc+ca=1\)

Ta có:

\(M=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)

\(=\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)\)

\(=\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)\left(c+a\right)\left(c+b\right)\)

\(=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{a+b+c}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{0}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

Lời giải:
$a+b+c=abc$

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

$\Leftrightarrow a^2+ab+ac+bc=bc(a^2+1)$

$\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\Leftrightarrow a^2+1=\frac{(a+b)(a+c)}{bc}$
Tương tự với $b^2+1, c^2+1$. Khi đó:

$Q=\frac{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}{bc.ac.ab}=[\frac{(a+b)(b+c)(c+a)}{abc}]^2$ là bình phương 1 số hữu tỉ.

Ta có đpcm.

Đặt \(\left(\frac{a}{b^2},\frac{b}{c^2},\frac{c}{a^2}\right)=\left(x,y,z\right)\)

\(\Rightarrow xyz=\frac{abc}{a^2b^2c^2}=\frac{1}{abc}=1\)

Theo bài ra ta có : \(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\)

\(\Leftrightarrow x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow x+y+z=xy+yz+xz\)

\(\Leftrightarrow\left(xy-x-y+1\right)-1+z\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(xy-x-y+1\right)+z\left(x+y-1-xy\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)-z\left(x-1\right)\left(y-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(1-z\right)=0\)

\(\Leftrightarrow\frac{a-b^2}{b^2}.\frac{b-c^2}{c^2}.\frac{a^2-c}{a^2}=0\)

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

Ta có đpcm

a + b + c = 0

=> (a + b + c)² = 0

=> a² + b² + c² + 2(ab + bc + ca) = 0

=> ab + bc + ca = \(\frac{a^2+b^2+c^2}{2}\)

=> \(\left(ab+bc+ca\right)^2=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)

=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2a^2bc+2ab^2c+2abc^2=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)

=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)

=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)(vì a + b + c = 0)

Lại có \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{a^2b^2+b^2c^2+a^2c^2}{a^2b^2c^2}=\frac{\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2}{\left(abc\right)^2}\)

\(=\frac{\left(\frac{a^2+b^2+c^2}{2}\right)^2}{\left(abc\right)^2}=\left(\frac{\frac{a^2+b^2+c^2}{2}}{abc}\right)^2=\left(\frac{a^2+b^2+c^2}{2abc}\right)^2\)

=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)là bình phương của 1 số hữu tỉ

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