a³+b³=(a+b)³-3ab(a+b)=1-3ab

Ta có \(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow3ab\le\frac{3}{4}\)

\(\Rightarrow1-3ab\ge\frac{1}{4}\)

Dấu "=" xảy ra khi a=b=\(\frac{1}{2}\)

Vậy..............

Ta có \(\frac{1}{a+b+1}=\left(1-\frac{1}{b+c+1}\right)+\left(1-\frac{1}{a+c+1}\right)=\frac{b+c}{b+c+1}+\frac{a+c}{a+c+1}\)

                                                                                                                   \(\ge2\sqrt{\frac{\left(b+c\right)\left(a+c\right)}{\left(b+c+1\right)\left(a+c+1\right)}}\)

Tương tự \(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{\left(a+b+1\right)\left(a+c+1\right)}}\)

                 \(\frac{1}{a+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)

Nhân 3 bđt trên ta có:

\(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(a+c\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\)

=> \(\left(a+b\right)\left(b+c\right)\left(a+c\right)\le\frac{1}{8}\)

MaxA=1/8 khi a=b=c=1/4

Ta có \(3=x+y+z=x+y+\frac{z}{2}+\frac{z}{2}\ge4\sqrt[4]{x.y.\frac{z^2}{4}}\)

=> \(xyz^2\le\frac{81}{64}\)

\(A=\frac{x+y}{xyz}\ge\frac{2\sqrt{xy}}{xyz}=\frac{2}{\sqrt{xyz^2}}\ge\frac{2}{\sqrt{\frac{81}{64}}}=\frac{16}{9}\)

MinA=16/9  khi \(x=y=\frac{3}{4};z=\frac{3}{2}\)

\(x=3-\sqrt{5}=\frac{1}{2}.\left(6-2\sqrt{5}\right)\)\(=\frac{1}{2}.\left(\sqrt{5}-1\right)^2\)

\(\Rightarrow x>0\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{1}{2}.\left(\sqrt{5}-1\right)^2}\)\(=\frac{\left|\sqrt{5}-1\right|}{\sqrt{2}}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{2}\)\(=\frac{\sqrt{10}-\sqrt{2}}{2}\)

Thay \(x=3-\sqrt{5};\sqrt{x}=\frac{\sqrt{10}-\sqrt{2}}{2}\)vào A ta được:

\(A=3-\sqrt{5}-2-\frac{\sqrt{10}-\sqrt{2}}{2}\)\(=\frac{6-2\sqrt{5}-4-\sqrt{10}+\sqrt{2}}{2}\)\(=\frac{2+\sqrt{2}-\sqrt{10}+2\sqrt{5}}{2}\)

\(x^2+\left(m-2\right)x-8=0\)

\(\Delta=b^2-4ac=\left(m-2\right)^2-4.1.\left(-8\right)=\left(m-2\right)^2+32\)

Vì \(\left(m-2\right)^2\ge0\forall m\)

\(\Rightarrow\left(m-2\right)^2+32\ge32>0\forall m\)

Vậy phương trình luôn có hai nghiệm phân biệt với mọi m

Theo định lí vi-ét ta có:\(\hept{\begin{cases}x_1+x_2=\frac{-b}{a}=2-m\\x_1x_2=\frac{c}{a}=-8\end{cases}}\Rightarrow x_2=\frac{-8}{x_1}\)

Theo bài ra ta có:\(A=\left(x_1^2-1\right)\left(x_2^2-4\right)=\left(x_1^2-1\right)\left(\frac{64}{x_1^2}-4\right)=68-4\left(x_1^2+\frac{16}{x_1^2}\right)\le68-4.8=36\)

Dấu "=" xảy ra <=> \(x_1=\pm2\)

+Với  \(x_1=2\Rightarrow m=4\)

+Với \(x_1=-2\Rightarrow m=0\)

Vậy \(A=\left(x_1^2-1\right)\left(x_2^2-4\right)\)đạt GTLN là 36 \(\Leftrightarrow m=0;m=4\)

Thay x =\(3-\sqrt{5}\) ta có:

\(A=3-\sqrt{5}-2-\sqrt{3-\sqrt{5}}\)

\(A=1-\sqrt{5}-\sqrt{3-\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{2}\left(1-\sqrt{5}\right)+\sqrt{5}-1\)

\(\sqrt{2}A=\left(1-\sqrt{5}\right)\left(\sqrt{2}-1\right)\)

\(A=\frac{\left(1-\sqrt{5}\right)\left(\sqrt{2}-1\right)}{\sqrt{2}}\)

Dòng thứ 3 nha bạn:

\(\sqrt{2}A=\sqrt{2}\left(1-\sqrt{5}\right)-\sqrt{6-2\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{2}\left(1-\sqrt{5}\right)-\sqrt{5-2\sqrt{5}+1}\)

\(\sqrt{2}A=\sqrt{2}\left(1-\sqrt{5}\right)-\sqrt{\left(\sqrt{5}-1\right)^2}\)

được chưa bạn???

đây nha bạn

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

\(\Rightarrow y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\)(1)

Tương tự \(x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\)(2)

Lấy (1) + (2) đc x + y = -x - y

                      <=> 2(x + y) = 0 

                      <=> x + y = 0