\(\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{b}\Leftrightarrow ab=bc+ac\Leftrightarrow2ab-2bc-2ac=0\\ \Leftrightarrow\sqrt{a^2+b^2+c^2}=\sqrt{a^2+b^2+c^2+2ab-2bc-2ac}\\ =\sqrt{\left(a+b-c\right)^2}=\left|a+b-c\right|\left(dpcm\right)\)

Câu 23:

https://olm.vn/hoi-dap/detail/1732532846797.html

Ta có: \(2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{2\left(a+b+c\right)}{abc}=0\)

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

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

Hằng đẳng thức:

\(\left(x-y-z\right)^2=x^2+y^2+z^2+2\left(yz-xy-zx\right)=x^2+y^2+z^2-2\left(xy+xz-yz\right)\)

\(\Rightarrow x^2+y^2+z^2=\left(x-y-z\right)^2+2\left(xy+xz-yz\right)\)

Giờ thay \(x=\dfrac{1}{a}\) ; \(y=\dfrac{1}{b}\); \(z=\dfrac{1}{c}\) là ra cái người ta làm

\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

\(=\sqrt{\dfrac{\left(bc\right)^2+\left(ac\right)^2+\left(ab\right)^2}{\left(abc\right)^2}}\)

\(=\dfrac{\sqrt{\left(bc+ac+ab\right)^2-2abc\left(a+b+c\right)}}{abc}\)

(áp dụng HĐT: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)\))

\(=\dfrac{\sqrt{\left[a\left(b+c\right)+bc\right]^2-2abc\left[a+\left(b+c\right)\right]}}{abc}\)

\(=\dfrac{\sqrt{\left(a^2+bc\right)^2-4a^2bc}}{abc}\)

\(=\dfrac{\sqrt{a^4+2a^2bc+\left(bc\right)^2-4a^2bc}}{abc}\)

\(=\dfrac{\sqrt{a^4-2a^2bc+\left(bc\right)^2}}{abc}\)

\(=\dfrac{a^2-bc}{abc}\) là 1 số hữu tỉ (đpcm)

Ta có:

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

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

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

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

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

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

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

Ta có : \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\text{=}\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2+2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

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

\(\text{=}\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2\left(do-a\text{=}b+c\right)\)

\(\Rightarrow\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\text{=}\sqrt{\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2}\)

\(\text{=}\left|\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right|\)

Do \(a,b,c\) là các số hữu tỉ khác 0 nên

\(\left|\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right|\) là một số hữu tỉ

\(\Rightarrow dpcm\)

Ta có : 

 P = \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\dfrac{1}{2ac}+\dfrac{1}{2ab}-\dfrac{1}{2bc}}\)

\(=\sqrt{\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\dfrac{1}{2abc}\left(b+c-a\right)}\)

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

=> P là số hữu tỉ với a,b,c \(\ne0\)

 P = 

 (do a = b + c) 

=> P là số hữu tỉ với a,b,c 

Ta có: \(a=b+c\Rightarrow a-b-c=0\)

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

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

Nên \(P=\sqrt[]{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt[]{\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2}\)

\(=\left|\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right|\) => ĐPCM

Bài này thiếu " a,b,c là các số hữu tỉ " phải không?

Lời giải:
\(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(a-c)^2}=(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a})^2-2(\frac{1}{(a-b)(b-c)}+\frac{1}{(b-c)(c-a)}+\frac{1}{(a-b)(c-a)})\)

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

\(\Rightarrow \sqrt{\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}}=|\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}|\) là số hữu tỷ (đpcm)