Question

[Trần Đan Khanh]

Câu 1 ,2 giải ntn ạ?😭😭 

Answer 1

[HT.Phong (9A5)]

3)

a) Ta có:

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

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

b) 

\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2024^2}+\dfrac{1}{2025^2}}\left(1\right)\\ =\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{\left(-3\right)^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{\left(-4\right)^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2024^2}+\dfrac{1}{\left(-2025\right)^2}}\)

Theo câu a \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) khi \(a+b+c=0\)

Mà: \(\left\{{}\begin{matrix}1+2+\left(-3\right)=0\\1+3+\left(-4\right)=0\\....\\1+2024+\left(-2025\right)=0\end{matrix}\right.\)  

=> \(\left(1\right)=\left|1+\dfrac{1}{2}+\dfrac{1}{-3}\right|+\left|1+\dfrac{1}{3}+\dfrac{1}{-4}\right|+...+\left|1+\dfrac{1}{2024}+\dfrac{1}{-2025}\right|\)  

Mà: \(\left\{{}\begin{matrix}1+\dfrac{1}{2}\cdot\dfrac{1}{-3}>0\\1+\dfrac{1}{3}+\dfrac{1}{-4}>0\\...\\1+\dfrac{1}{2024}+\dfrac{1}{-2025}>0\end{matrix}\right.\) 

=> \(\left(1\right)=1+\dfrac{1}{2}+\dfrac{1}{-3}+1+\dfrac{1}{3}+\dfrac{1}{-4}+...+1+\dfrac{1}{2024}+\dfrac{1}{-2025}\\ =\left(1+1+...+1\right)+\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2024}-\dfrac{1}{2024}\right)-\dfrac{1}{2025}\\ =2023+\dfrac{1}{2}-\dfrac{1}{2025}\)

Answer 2

[HT.Phong (9A5)]

4)

\(\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{14-8\sqrt{3}}}-3\sqrt{2\left(1-\sqrt{-2\sqrt{3}+4}\right)+2\sqrt{4+2\sqrt{3}}}\\ =\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{4-2\sqrt{3}}\right)+2\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}\\ =\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{\left(2\sqrt{2}-\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}\right)+2\sqrt{\left(\sqrt{3}+1\right)^2}}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}-1\right)^2}\right)+2\left(\sqrt{3}+1\right)}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2\left(1-\sqrt{3}+1\right)+2\sqrt{3}+2}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2-2\sqrt{3}+2+2\sqrt{3}+2}\\ =3\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{6}\right)\cdot\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{6}\\ =3\sqrt{2}\cdot\sqrt{3}-3\sqrt{6}\\ =3\sqrt{6}-3\sqrt{6}\\ =0\)