\(S=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

\(\frac{\sqrt{2}}{\sqrt{3}}S=\frac{\sqrt{2}}{\sqrt{3}}\sqrt{a+b}+\frac{\sqrt{2}}{\sqrt{3}}\sqrt{b+c}+\frac{\sqrt{2}}{\sqrt{3}}\sqrt{c+a}\)

\(\le\frac{\frac{2}{3}+a+b}{2}+\frac{\frac{2}{3}+b+c}{2}+\frac{\frac{2}{3}+c+a}{2}\)

\(=1+a+b+c=2\)

\(\Rightarrow S\le\frac{2}{\frac{\sqrt{2}}{\sqrt{3}}}=\sqrt{6}\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

áp dụng bất đẳng thức Cô-si ta có:

\(\left(a+b\right)+\frac{2}{3}\ge2\sqrt{\frac{2}{3}\left(a+b\right)}=\sqrt{\frac{8}{3}}.\sqrt{a+b}\)

\(\left(b+c\right)+\frac{2}{3}\ge2\sqrt{\frac{2}{3}\left(b+c\right)}=\sqrt{\frac{8}{3}}.\sqrt{b+c}\)

\(\left(c+a\right)+\frac{2}{3}\ge2\sqrt{\frac{2}{3}.\left(c+a\right)}=\sqrt{\frac{8}{3}}.\sqrt{c+a}\)

\(\Rightarrow2\left(a+b+c\right)+2\ge\sqrt{\frac{8}{3}}.\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\)

\(\Rightarrow4\ge\sqrt{\frac{8}{3}}.S\Leftrightarrow S\le\sqrt{6}\)

dấu bằng xảy ra khi a=b=c

hình như đề sai hay sao ấy

tách mãi mà vẫn cứ phụ thuộc

đặt \(\sin\left(a\right)^2=x;\cos\left(a\right)^2=y;x+y=1\)

Ta có:

\(N=\sqrt{x^2+4y+\sqrt{y^2+4x}}=\sqrt{x^2+4\left(1-x\right)+\sqrt{y^2-4\left(1-y\right)}}\)

\(=\sqrt{x^2-4x+4+\sqrt{y^2-4y+4}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(y-2\right)^2}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(1-x-2\right)^2}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(x+1\right)^2}}\)\(=\sqrt{x^2-4x+4+x+1}=\sqrt{x^2-3x+5}\)

để x²+x+1991 là số chính phương

=>x²+x là stn

=>x là số nguyên

đặt x²+x+1991=a²

=>4x²+4x+1991.4=4a²

=>(2x+1)²+7963=4a²

=>(2a-2x-1)(2a+2x+1)=7963

từ đó tìm x là được

x hữu tỷ mà

trang cho oi dau bdt sai roi ^.^ 

tao giai cho may ne 

\(\frac{3+a}{3-a}+\frac{3+b}{3-b}+\frac{3+c}{3-c}\)=\(\frac{2a+b+c}{b+c}+\frac{2b+a+c}{a+c}+\frac{2c+a+b}{a+b}\)

=\(2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+3\ge2\cdot\frac{3}{2}+3=6\)

đến đây tự làm nhé

BĐT Sai kìa.

(a,b,c)=(1;0.5;1,5) 
 

\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

   \(=\sqrt{\frac{2\left(4+\sqrt{7}\right)}{2}}-\sqrt{\frac{2\left(4-\sqrt{7}\right)}{2}}\)

   \(=\sqrt{\frac{8+2\sqrt{7}}{2}}-\sqrt{\frac{8-2\sqrt{7}}{2}}\)

   \(=\sqrt{\frac{7+2\sqrt{7}+1}{2}}-\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)

   \(=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}\)

   \(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)

   \(=\frac{|\sqrt{7}+1|}{\sqrt{2}}-\frac{|\sqrt{7}-1|}{\sqrt{2}}\)

   \(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}\)

   \(=\frac{2}{\sqrt{2}}\)

cho mình hỏi tại sao chia 2 vậy?

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-1=9\)

rtrrfgfffjhfhfd

\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

\(\Leftrightarrow x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3\sqrt[3]{\sqrt[3]{\frac{1}{4-\sqrt{5}}}.\sqrt[3]{4-\sqrt{5}}}.x\)

\(=4+\sqrt{15}+4-\sqrt{15}+3x=8+3x\)

=>y=3x+8-3x+1987

=1995