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Thay ab+bc+ac = 1 và Q ta được :
\(Q=\left(a^2+ab+ac+bc\right)\left(b^2+ab+ac+bc\right)\left(c^2+ab+ac+bc\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) là bình phương của một số hữu tỉ (đpcm)
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:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)
\(\Leftrightarrow\frac{ab+ac+bc}{abc}=\frac{1}{abc}\)
\(\Rightarrow ab+ac+bc=1\)
Ta có :
\(1+a^2=ab+ac+bc+a^2=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)
\(1+b^2=ab+ac+bc+b^2=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\)
\(1+c^2=ab+ac+bc+c^2=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) là bình phương của 1 số hữu tỉ (ĐPCM)
Đặt x = a - b ; y = b - c ; z = c - a thì x + y + z = a - b + b - c + c - a = 0
Ta có \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}\)
= \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
= \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\frac{x+y+z}{xyz}\)
= \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2\)( đpcm )
+ \(a^2+1=a^2+ab+bc+ca\)
\(=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
+ Tương từ ta cm đc :
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Do đó : \(Q=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow Q=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Vì a,b,c là các số hữu tỉ nên \(\left(s+b\right)\left(b+c\right)\left(c+a\right)\)là số hữu tỉ
Do đó suy ra đpcm
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow ab+bc+ca=1\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)\(=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)
\(=\left(a+c\right)\left(b+a\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow\frac{bc+ac+ab}{abc}=\frac{1}{abc}\left(QĐ\right)\Leftrightarrow ac+bc+ab=1\)
\(\Rightarrow1+a^2=bc+ab+ac+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+c\right)\left(a+b\right)\)
Tương tự: \(1+b^2=\left(a+b\right)\left(b+c\right)\); \(1+c^2=\left(a+c\right)\left(b+c\right)\)
Nhân vế với vế ta được: \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)
mà \(\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)là số chính phương => đpcm