mk ko bt viết sigma trên đây :'< bn thông cảm

Đặt \(A=\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)

\(=\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{a+b+c}-4\)

\(=\left(a+b+c+d\right)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}\right)-4\)

\(\ge\frac{16\left(a+b+c+d\right)}{3\left(a+b+c+d\right)}-4=\frac{16}{3}-4=\frac{4}{3}\)

Đặt \(B=\frac{b+c+d}{a}+\frac{a+c+d}{b}+\frac{a+b+d}{c}+\frac{a+b+c}{d}\)

\(=\frac{a+b+c+d}{a}+\frac{a+b+c+d}{b}+\frac{a+b+c+d}{c}+\frac{a+b+c+d}{d}-4\)

\(=\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)-4\ge\frac{16\left(a+b+c+d\right)}{a+b+c+d}-4=12\)

\(\Rightarrow\)\(S=A+B\ge\frac{4}{3}+12=\frac{40}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=d\)

\(\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(\frac{1}{y}+y\right)=\frac{9}{2}\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=5\end{cases}}\)

dat an phu r giai 

AB=căn 4,5^2+6^2=7.5

B, AH= căn 6^2-3^2=3 căn 3

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

\(=\frac{\sqrt{4}}{3+\sqrt{5}}+\frac{\sqrt{4}}{3-\sqrt{5}}=\frac{2.\left(3-\sqrt{5}\right)+2.\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\frac{12}{4}=3\)

\(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\)  + \(\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}\)

= \(\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{9-5}}\)+ \(\sqrt{\frac{\left(3+\sqrt{5}\right)^2}{9-5}}\)

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

= \(\frac{3-\sqrt{5}}{2}\)+ \(\frac{3+\sqrt{5}}{2}\)

= \(\frac{6}{2}\)

=3

#mã mã#

ĐK \(x\ge-\frac{2}{3}\)

Pt

<=> \(x^3+2x^2-4x-3+3\left(x+1\right)\left(x+1-\sqrt{3x+2}\right)=0\)

<=> \(\left(x+3\right)\left(x^2-x-1\right)+3\left(x+1\right).\frac{\left(x+1\right)^2-3x-2}{x+1+\sqrt{3x+2}}=0\)

<=> \(\left(x+3\right)\left(x^2-x-1\right)+3\left(x+1\right).\frac{x^2-x-1}{x+1+\sqrt{3x+2}}=0\)

<=> \(\orbr{\begin{cases}x^2-x-1=0\\x+3+\frac{3\left(x+1\right)}{x+1+\sqrt{3x+2}}=0\left(2\right)\end{cases}}\)

Pt (2) vô nghiệm do VT>0 với mọi \(x\ge-\frac{2}{3}\)

=> \(x=\frac{1\pm\sqrt{5}}{2}\)(tmĐKXĐ)

Vậy \(x=\frac{1\pm\sqrt{5}}{2}\)