mình ko biết

\(M=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\frac{a+b+c}{abc}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\) là số hữu tỉ

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

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

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

Do a,b,c là các số hữu tỉ => đpcm

Ta có 

\(\frac{1}{a^2\:}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}-\frac{1}{b\:}-\frac{1}{c}\right)^2\)².    + \(2\left(\frac{1}{ab}+\frac{1}{ac}-\frac{1}{bc}\right)\)

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

Từ đó suy ra 

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)\(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2}\)\(=|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}|\)Vì a,b,c là số hữu tỉ khác 0 nên \(|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}|\)là một số hữu tỉ

=> đpcm

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không hiện link thì mình gửi qua tin nhắn nhé

Ta có: ab2+bc2+ca2=a2c+b2a+c2bab2+bc2+ca2=a2c+b2a+c2b

⇔a3c2+b3a2+c3b2=b3c+c3a+a3b

⇔a3c2+b3a2+c3b2=b3c+c3a+a3b ( Do a2b2c2=abc=1)

⇔ a3c2+b3a2+c3b2 -b3c-c3a-a3b+a2b2c2-abc=0( Do a2b2c2=abc=1)

⇔(a2b2c2−a3c2)−(b3a2−a3b)−(c3b2−c3a)+(b3c−abc)=0

⇔(a2b2c2−a3c2)−(b3a2−a3b)−(c3b2−c3a)+(b3c−abc)=0

Tự phân tích thành nhân tử nhá: ⇔(b2−a)(c2−b)(a2−c)=0⇔(b2−a)(c2−b)(a2−c)=0

Đến đây suy ra ĐPCM

Ta có:

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

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

\(=\frac{b^4+2b^3c+3b^2c^2+2bc^3+c^4}{b^2c^2\left(b+c\right)^2}\)

\(=\frac{\left(b^4+2b^2c^2+c^4\right)+2bc\left(b^2+c^2\right)+b^2c^2}{b^2c^2\left(b+c\right)^2}\)

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

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

Vì a, b, c là các số hữu tỷ nên \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\) là số hữu tỷ

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