\(A=\frac{10a^2+10b^2+c^2}{ab+bc+ca}=\frac{8a^2+\frac{c^2}{2}+8b^2+\frac{c^2}{2}+2a^2+2b^2}{ab+bc+ca}\)

\(\ge\frac{2\sqrt{8a^2.\frac{c^2}{2}}+2\sqrt{8b^2.\frac{c^2}{2}}+4\sqrt{a^2b^2}}{ab+bc+ca}=\frac{4\left(ab+bc+ca\right)}{ab+bc+ca}=4\)

Dấu \(=\)khi \(a=b=\frac{c}{4}\).

Bạn tham khảo nhé: áp dụng bđt côsi cho 2 số dương

2a²+2b²>=4ab;8a²+c²/2>=4ac;8b²+c²/2>=4ac nên A>=4

dấu bằng xảy ra khi 4a=4b=c

\(D=\sqrt{9-\sqrt{87}}\sqrt{9+\sqrt{87}}=\sqrt{81-87}\)

đề sai ko bạn vì \(\sqrt{a}\)xảy ra khi a >= 0 mà -6 < 0 bạn nhé 

Ta có 

\(P=\frac{\sqrt{x}-1}{\sqrt{x}}>0\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)vì \(\sqrt{x}\ge0\)

Ta có:

\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{\left(1+n+n^2\right)^2}{n^2\left(n+1\right)^2}}\)

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

Áp dụng bài toán được

\(A=\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{2021^2}+\frac{1}{2022^2}}\)

\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}+\frac{1}{4}+...+1+\frac{1}{2021}-\frac{1}{2022}\)

\(=2020+\frac{1}{2}-\frac{1}{2022}=\)