Đặt \(\sqrt{x}=a\left(a\ge0\right)\)

\(M=\frac{a+1}{a^2+a+1}=\frac{a^2+a+1-a^2}{a^2+a+1}=1-\frac{a^2}{a^2+a+1}\le1\)

"=" xảy ra <=> a = 0 hay x = 0 

Tham khảo:

a + b + c = 0

b² + c² - a² = (b + c)² - a² - 2bc = (a + b + c)(b + c - a) - 2bc = - 2bc

c² + a² - b² = (c + a)² - b² - 2ca = (a + b + c)(c + a - b) - 2ca = - 2ca

a² + b² - c² = (a + b)² - c² - 2ab = (a + b + c)(a + b - c) - 2ab = - 2ab

A = 1/(b² + c² - a²) + 1/(c² + a² - b² ) + 1/(a² + b² - c²)

= - (1/2)(1/bc + 1/ca + 1/ab)

= - (1/2)(a + b + c)/abc = 0

K mình nha

mondora nài hăm bít

a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)

\(\sqrt{6}< \sqrt{6,25}=2,5\); 

\(\sqrt{12}< \sqrt{12,25}=3,5\); 

\(\sqrt{20}< \sqrt{20,25}=4,5\)

=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)

Vậy P < 12

Answer:

ý a, tham khảo bài làm của @xyzquynhdi

\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)

\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)