\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{a^2\left(a+1\right)^2+a^2+\left(a+1\right)^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)

\(=\sqrt{\dfrac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)

\(=\sqrt{\dfrac{\left(a\left(a+1\right)+1\right)^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\dfrac{a\left(a+1\right)+1}{a+1}+\dfrac{a}{a+1}\)

\(=\dfrac{a^2+2a+1}{a+1}=\dfrac{\left(a+1\right)^2}{a+1}=a+1\)

\(\Rightarrow VP=6831\)

Không làm mất tính tổng quát, giả sử \(x\le y\le z\)

Dễ dàng kiểm chứng \(x=y=z\) không phải là nghiệm

\(3^x+3^y+3^z=6831\Leftrightarrow3^x\left(1+3^{y-x}+3^{z-x}\right)=3^3.253\)

Nếu \(1+3^{y-x}+3^{z-x}\ne253\Rightarrow1+3^{y-x}+3^{z-x}=253.3^k⋮3\)

Nhưng \(1+3^{y-x}+3^{z-x}⋮̸3\) với \(\left\{{}\begin{matrix}x\ne y\\x\ne z\end{matrix}\right.\)\(\Rightarrow\) vô lý

Vậy \(\left\{{}\begin{matrix}3^x=3^3\\1+3^{y-x}+3^{z-x}=253\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\3^{y-3}+3^{z-3}=252\end{matrix}\right.\)

\(\Rightarrow3^{y-3}\left(1+3^{z-y}\right)=252=3^2.28\)

Do \(3^{z-y}+1⋮̸3\) lý luậnt ương tự như trên \(\Rightarrow\left\{{}\begin{matrix}3^{y-3}=3^2\\1+3^{z-y}=28\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y-3=2\\3^{z-y}=27=3^3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=5\\z=8\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=3\\y=5\\z=8\end{matrix}\right.\)

Đặt \(a=\sqrt{2x-3}\) ; \(b=\sqrt{y-2}\) ; \(c=\sqrt{3z-1}\) (\(a,b,c>0\))

Ta có : \(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c=14\)

\(\Leftrightarrow\left(\sqrt{2x-3}+\frac{1}{\sqrt{2x-3}}-2\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{3z-1}+\frac{16}{\sqrt{3z-1}}-8\right)=0\)

\(\Leftrightarrow\left[\frac{\left(2x-3\right)-2\sqrt{2x-3}+1}{\sqrt{2x-3}}\right]+\left[\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}\right]+\left[\frac{\left(3z-1\right)-8\sqrt{3z-1}+16}{\sqrt{3z-1}}\right]=0\)

\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)^2}{\sqrt{2x-3}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{3z-1}-4\right)^2}{\sqrt{3z-1}}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2x-3}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{3z-1}-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}}\)(TMĐK)

Vậy : \(\left(x;y;z\right)=\left(2;6;\frac{17}{3}\right)\)

Phần đặt ẩn a,b,c bạn bỏ đi nhé ^^

\(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)

Trường hợp x=y=z thì không phải bàn,ns cái trường hợp x+y+z=0

\(\frac{1}{x^2+y^2-z^2}=\frac{1}{\left(x+y\right)^2-2xy-z^2}=\frac{1}{\left(-z\right)^2-z^2-2xy}=\frac{1}{-2xy}\)

Tương tự rồi cộng lại thì \(BT=0\) thì phải

Condition\(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)

Put \(P=\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)

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

Because \(x^2+y^2+z^2=3xyz\)

\(\Leftrightarrow x^2+y^2+z^2-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)ư\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{cases}}\)

The first case: If \(x+y+z=0\left(1\right)\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}\left(2\right)}\)

From \(\left(1\right)\Rightarrow\hept{\begin{cases}x-y=-2y-z\\y-z=-2z-x\\z-x=-2x-y\end{cases}\left(3\right)}\)

 \(\left(2\right)\)and \(\left(3\right)\)into \(\left(4\right)\)we have

\(P=\frac{1}{x^2-x\left(-2z-x\right)}+\frac{1}{y^2-y\left(-2x-y\right)}+\frac{1}{z^2-z\left(-2y-z\right)}\)

\(=\frac{1}{2x^2+2xz}+\frac{1}{2y^2+2xy}+\frac{1}{2z^2+2yz}\)

\(=\frac{1}{2x\left(x+z\right)}+\frac{1}{2y\left(x+y\right)}+\frac{1}{2z\left(z+y\right)}\)

\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)

\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)

\(=\frac{z+x+y}{-2xyz}=0\)( Because x+y+z=0)

The second case:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\left(5\right)\)

We have \(\hept{\begin{cases}\left(x-y\right)^2\ge0;\forall x,y,z\\\left(y-z\right)^2\ge0;\forall x,y,z\\\left(z-x\right)^2\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0;\forall x,y,z\left(6\right)\)

From \(\left(5\right),\left(6\right)\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)

Because \(x=y=z\Rightarrow x^2=y^2=z^2=xy=yz=zx\)

So \(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

\(=\frac{z+x+y}{xyz}=0\)

So...

Ta có hđt \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow x^3+y^3+z^3=3xyz\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)

Nếu x + y + z = 0 thì bt = 0

Nếu x = y = z thì bt = 3 

bài này dễ nhưng bạn phải chứng minh bđt này đã:

\(\frac{1}{a+b+c+d}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\)

với a;b;c;d là các số dương

bạn có thể cm bđt trên bằng cách biến đổi tương đương hoặc cm bđt Schwat (Sơ-vác)

Mình là 1 phần tử đại diện còn lại là hoàn toàn tt nhé 

ta có \(\frac{1}{3\sqrt{x}+3\sqrt{y}+2\sqrt{z}}=\frac{1}{2\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{y}+\sqrt{z}\right)+\left(\sqrt{x}+\sqrt{z}\right)}\)

\(\le\frac{1}{16}\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{x}+\sqrt{z}}\right)\)

Tương tự ta cm được 

\(VT\le\frac{1}{16}.4\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{z}+\sqrt{x}}\right)\)\(=\frac{1}{4}.3=\frac{3}{4}\)

dấu "=" khi x=y=z