Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=4\) (*)

Mà \(a+b+c=abc\Rightarrow\frac{a+b+c}{abc}=1\)

Từ (*) \(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\Rightarrowđpcm\)

Đặt \(x=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Rightarrow x^2=8+2\sqrt{6-2\sqrt{5}}=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)

\(\Rightarrow x=\sqrt{5}+1\)

Vậy A=1

a) \(p^2q+p⋮\left(p^2+q\right)\Rightarrow q\left(p^2+q\right)-\left(p^2q+q\right)=q^2-p\left(p^2+q\right)\)

\(pq^2+q⋮\left(q^2-p\right)\Rightarrow\left(pq^2+q\right)-p\left(q^2-p\right)=p^2+q⋮q^2-p\)

\(q^2-p=-\left(p^2+q\right)\Leftrightarrow q^2+q+p^2-p=0\left(VN\right)\)

\(q^2-p=p^2+q\Leftrightarrow\left(q+p\right)\left(q-p-1\right)=0\Leftrightarrow q-p-1=0\Leftrightarrow q=p+1\)

Mà p,q là 2 số nguyên tố nên p=2, q=3 

\(pt\Leftrightarrow2x+\sqrt{\left(\sqrt{x-\frac{1}{4}+\frac{1}{2}}\right)^2}=2\)

đk: \(x\ge\frac{1}{4}\)

\(\Leftrightarrow2x+\left(\sqrt{x-\frac{1}{4}}+\frac{1}{2}\right)=2\Leftrightarrow\sqrt{x-\frac{1}{4}}=\frac{3}{2}-2x\)

\(\Leftrightarrow\hept{\begin{cases}\frac{3}{2}-2x\ge0\\x-\frac{1}{4}=\left(\frac{3}{2}-2x\right)\end{cases}^2\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{4}\\x-\frac{1}{4}=\frac{9}{4}-6x+4x^2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{4}\\x=\frac{1}{2}\\x=\frac{5}{4}\left(loại\right)\end{cases}}\)

Vậy S={1/2}