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: \(a=b+c\Rightarrow c=a-b\)

\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\dfrac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{b^2\left(a-b\right)^2+a^2\left(a-b\right)^2+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{b^4+a^2b^2-2ab^3+a^4+a^2b^2-2a^3b+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{\left(a^2+b^2\right)^2-2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{\left(a^2+b^2-ab\right)^2}{a^2b^2c^2}}=\left|\dfrac{a^2+b^2-ab}{abc}\right|\)

=> Là một số hữu tỉ do a,b,c là số hữu tỉ

a) Từ giả thiết : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)

\(\Rightarrow2ab\text{=}2bc+2ca\)

\(\Rightarrow2ab-2bc-2ca\text{=}0\)

Ta xét : \(\left(a+b-c\right)^2\text{=}a^2+b^2+c^2+2ab-2bc-2ca\)

\(\text{=}a^2+b^2+c^2\)

Do đó : \(A\text{=}\sqrt{a^2+b^2+c^2}\text{=}\sqrt{\left(a+b-c\right)^2}\)

\(\Rightarrow A\text{=}a+b-c\)

Vì a;b;c là các số hữu tỉ suy ra : đpcm

b) Đặt : \(a\text{=}\dfrac{1}{x-y};b\text{=}\dfrac{1}{y-x};c\text{=}\dfrac{1}{z-x}\)

Do đó : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)

Ta có : \(B\text{=}\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

Từ đây ta thấy giống phần a nên :

\(B\text{=}a+b-c\)

\(B\text{=}\dfrac{1}{x-y}+\dfrac{1}{y-z}-\dfrac{1}{z-x}\)

Suy ra : đpcm.

Mình bổ sung đề phần b cần phải có điều kiện của x;y;z nha bạn.

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

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

\(=\sqrt{\dfrac{b^4+a^4-2ab^3-2a^3b+3a^2b^2}{a^2b^2\left(a-b\right)^2}}=\sqrt{\dfrac{\left(b^2+a^2\right)^2-2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)

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

Do a,b là số hữu tỉ\(\Rightarrow\)\(\left|\dfrac{a^2+b^2-ab}{ab\left(a-b\right)}\right|\) là số hữu tỉ hay A là số hữu tỉ

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{a+b+c}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{0}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\sqrt{\dfrac{1}{4}+\dfrac{1}{\left(2n-1\right)^2}+\dfrac{1}{\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(2n-1\right)^2\left(2n+1\right)^2+4\left(2n-1\right)^2+4\left(2n+1\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(4n^2-1\right)^2+4\left(4n^2-4n+1\right)+4\left(4n^2+4n+1\right)}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)

\(=\sqrt{\dfrac{16n^4+24n^2+9}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(4n^2+3\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\dfrac{4n^2+3}{2\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{\left(4n^2-1\right)+4}{2\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{1}{2}+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

Do đó:

\(P=\left(\dfrac{1}{2}+\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{399}-\dfrac{1}{401}\right)\)

\(=\dfrac{1}{2}.200+1-\dfrac{1}{401}=\dfrac{40500}{401}\)

\(\Rightarrow Q=400\)

Câu 1:

Thay \(x=-12\) vào \(\left|x-2\right|\)

\(\Rightarrow\left|-12-2\right|=\left|-14\right|=14\)

Câu 2: Chọn phương án A.

Câu 3:

\(\left|-120\right|+\left|20\right|=120+20=140\)

Câu `1`

` |x + 2|`

mà `x=-12`

`->  |-12 + 2|= |-10|=10`

`->B`

Câu `2`

`->A`

Câu `3`

`A = |-120| + |20|`

`= 120 +20`

`=140`