Thay\(x=\dfrac{16}{9}\)vào A ta đc

A=\(\dfrac{\sqrt{\dfrac{16}{9}}+1}{\sqrt{\dfrac{16}{9}}-1}=\dfrac{\dfrac{4}{3}+1}{\dfrac{4}{3}-1}=\dfrac{21}{3}=7\)

Thay x=\(\dfrac{25}{9}\)vào A ta đươc

A=\(\dfrac{\sqrt{\dfrac{25}{9}}+1}{\sqrt{\dfrac{25}{9}-1}}=\dfrac{\dfrac{5}{3}+1}{\dfrac{5}{3}-1}=\dfrac{8}{2}=4\)

Vậy....................

Để\(\sqrt{x^2-4}+\sqrt{x-2}=0\) thì \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\x=2\end{matrix}\right.\Leftrightarrow x=2\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Leftrightarrow\dfrac{a}{b}=1\Leftrightarrow a=b\)) (1)

\(\dfrac{b}{c}=1\Leftrightarrow b=c\) (2)

\(\dfrac{c}{a}=1\Leftrightarrow c=a\) (3)

Từ (1) , (2) và (3) thì a=b=c (đpcm)

Đề thiếu bn ơi

\(D=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=1-\dfrac{1}{6}=\dfrac{5}{6}\)

Bn chỉ cần lám\(\dfrac{7}{3n-1}nguyênthôirôiitìmn\)