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

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

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

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

Do a,b,c là các số hữu tỉ khác 0 nên \(|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}|\) là 1 số hữu tỉ

=.= hok tốt!!

Ta có: 

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

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

Suy ra: 

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

Do a, b, c là các số hữu tỉ khác 0 nên \(\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\)là một số hữu tỉ.

(Chúc bạn làm bài tốt và nhớ click cho mình với nhá!)

Đặt \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) có \(a=b+c\Rightarrow A=\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{b^2c^2+c^2\left(b+c\right)^2+b^2\left(b+c\right)^2}{\left(b+c\right)^2b^2c^2}\)
Ta có \(b^2c^2+c^2\left(b+c\right)^2+b^2\left(b+c\right)^2=b^2c^2+\left(b+c\right)^2\left(b^2+c^2\right)\)
        =\(b^2c^2+\left(b^2+c^2+2bc\right)\left(b^2+c^2\right)=b^2c^2+\left(b^2+c^2\right)^2+2bc\left(b^2+c^2\right)\)
        =\(\left(bc+\left(b^2+c^2\right)\right)^2\)
Vậy \(A=\frac{\left(bc+\left(b^2+c^2\right)\right)^2}{\left(b+c\right)^2b^2c^2}\Rightarrow\sqrt{A}=\frac{bc+b^2+c^2}{\left|\left(b+c\right)bc\right|}\)
Do \(b,c\)là các số chính phương nên \(\sqrt{A}\)chính phương suy ra điều phải chứng minh.

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

= \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\)= A²

vậy A = \(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)là một số hữu tỉ

Câu hỏi của Phạm Quang Dương - Toán lớp 9 - Học toán với OnlineMath

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

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

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

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

Vậy \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)là số hữu tỉ

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

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

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

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

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

Lời giải:
Bạn chú ý lần sau gõ đề bài bằng công thức toán. Việc gõ đề thiếu/ sai/ không đúng công thức khiến người sửa rất mệt.

a) Theo hằng đẳng thức đáng nhớ:

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

\(\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\frac{2(a+b+c)}{abc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-0}\) (do $a+b+c=0$)

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

b) Theo điều kiện đề bài:

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

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

Vì \(a,b,c\in\mathbb{Q}\Rightarrow \)\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{b+c}-\frac{b+c}{bc}\right|\in\mathbb{Q}\)

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

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

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

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