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 \(n=2\).

uk khỏi cần

k nha

đc rồi khỏi trả lời nha

ĐK: \(x\ne-1,y\ne2\).

\(\hept{\begin{cases}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{x+1}+\frac{2}{y-2}=5\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\)

Đặt \(\frac{1}{x+1}=a,\frac{1}{y-2}=b\).

Hệ phương trình đã cho trở thành: 

\(\hept{\begin{cases}a+2b=5\\5a-b=3\end{cases}}\Leftrightarrow\hept{\begin{cases}b=5a-3\\a+2\left(5a-3\right)=5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{x+1}=1\\\frac{1}{y-2}=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=\frac{5}{2}\end{cases}}\left(tm\right)\)

\(\frac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right)x^2-\sqrt{6}}=\frac{x^2-\sqrt{2}}{x^4+\sqrt{3}x^2-\sqrt{2}x^2-\sqrt{6}}\)

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

\(\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)

\(=\left(12-2.2\sqrt{2}+1\right)\left(4+4\sqrt{3}+3\right)-8\sqrt{20+2\sqrt{27+2.4.3\sqrt{3}+16}}\)

\(=\left(2\sqrt{2}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)

\(=\left(2\sqrt{2}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+6\sqrt{3}+8}\)

\(=\left(2\sqrt{2}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{27+2.3\sqrt{3}+1}\)

\(=\left(2\sqrt{2}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)

Đến đây thì cũng chưa biết phải làm gì luôn =))