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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)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)
\(\Leftrightarrow\frac{ab+ac+bc}{abc}=\frac{1}{abc}\)
\(\Rightarrow ab+ac+bc=1\)
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)
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:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(A=a^5-a=a.\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)=a\left(a-1\right)\left(a+1\right)\left(a^2+1\right)=B\left(a^2+1\right)\)B là 3 số tự nhiên liên tiếp \(\left\{{}\begin{matrix}B⋮2\\B⋮3\\B⋮6\end{matrix}\right.\) ta cần c/m A chia cho 5
\(A=B\left(n^2+1\right)=B\left[\left(n^2-4\right)+5\right]=B\left(n^2-2^2\right)=B\left(n-2\right)\left(n+2\right)+5B=C+5B\)C là tích 5 số tự nhiên liên tiếp: \(\left\{{}\begin{matrix}C⋮5\\5B⋮5\end{matrix}\right.\)\(\Rightarrow A⋮5\)
\(\left\{{}\begin{matrix}A⋮5\\A⋮6\end{matrix}\right.\)\(\Rightarrow A⋮30\) => dpcm
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 )
- Cái này mình tham khảo chứ bó tay rồi :)
* Đặt x=a-b ; y=b-c ; z=c-a thì x+y+=a-b+b-c+c-a=0
* \(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\)=\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)=\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2.\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)\)=\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2.\left(\dfrac{x+y+z}{xyz}\right)\)=\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\)=\(\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)^2\)