=  \(\frac{\sqrt{5}}{\sqrt{\sqrt{5}+1}+1}\) + \(\frac{\sqrt{5}}{\sqrt{\sqrt{5}+1}-1}\)

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

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

= \(2\sqrt{\sqrt{5}+1}\)

z² nha

\(x=\frac{a}{b}=\frac{a+b}{a}\)

xét \(\frac{a}{b}=\frac{a+b}{a}\)

\(< =>a^2=ab+b^2\)

\(a^2-ab-b^2=0\)

\(\frac{a^2}{b^2}-\frac{a}{b}-1=0\)

\(\left(\frac{a}{b}\right)^2-\frac{a}{b}-1=0\)

đặt \(\frac{a}{b}=c\)

\(c^2-c-1=0\)

\(a=1;b=-1;c=-1\)

\(\Delta=\left(-1\right)^2-\left(4.1.-1\right)=1+4=5\)

\(\sqrt{\Delta}=\sqrt{5}\)

\(c_1=\frac{1+\sqrt{5}}{2}\left(TM\right)\)

\(c_2=\frac{1-\sqrt{5}}{2}\left(KTM\right)\)kết hợp đkxđ: \(a,b>0\)

mà \(1-\sqrt{5}< 0\left(KTM\right)\)

\(< =>\frac{a}{b}=\frac{1+\sqrt{5}}{2}=x\)

\(x=\frac{1+\sqrt{5}}{2}\)

Ta có:A = \(x^3+y^3+z^3=\left(x+y\right)^3+z^3-3xy\left(x+y\right)\)

A = \(\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz\right)-3xy\left(x+y+z\right)+3xyz\)

A = \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)

A = \(3\left(x^2+y^2+z^2-xy-yz-xz+xyz\right)\) (Vì x + y + z = 3)

Mà: \(0\le x,y,z\le1\) => \(\hept{\begin{cases}x-1\le0\\y-1\le0\\z-1\le0\end{cases}}\) => (x - 1)(y - 1)(z - 1) \(\le\)0 => xyz - (xy + yz + xz) + (x + y + z) - 1\(\le\)0

<=> xyz \(\le\)xy + yz + xz + 1 - 3  = xy + yz + xz  - 2

Do đó: A \(\le3\left(x^2+y^2+z^2-xy-yz-xz+xy+yz+xz-2\right)\)

A \(\le3\left(x^2+y^2+z^2-2\right)\le3\cdot\left(\frac{\left(x+y+z\right)^2}{3}-2\right)=3\cdot\left(\frac{3^2}{3}-2\right)=3\)

Dấu "=" xảy ra<=> x = y = z = 1

Vậy Max x³ + y³ + z³ = 3 <=> x = y = z = 1

thanks bạn nhìu nha

\(A=\sqrt{\left(3+2\sqrt{2}\right)^n}+\sqrt{\left(3-2\sqrt{2}\right)^n}\)

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

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

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

Với \(n=1\):

\(A=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\ne6\)

Với \(n=2\):

\(A=3+2\sqrt{2}+3-2\sqrt{2}=6\)(thỏa) 

Với \(n\ge3\):

\(A>\left(\sqrt{2}+1\right)^3=5\sqrt{2}+7>6\)

Do đó \(n\)nguyên dương cần tìm là \(n=2\).