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

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

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

2,

\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)

\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)

\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)

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

Ta có \(\frac{m\sqrt{n}-n\sqrt{m}}{m-n}=\frac{\sqrt{mn}\left(\sqrt{m}-\sqrt{n}\right)}{\left(\sqrt{m}-\sqrt{n}\right)\left(\sqrt{m}+\sqrt{n}\right)}=\frac{\sqrt{mn}}{\sqrt{m}+\sqrt{n}}\)

Bạn tìm GTNN theo z thì đề đúng bằng cách:

(x+y)(1/x+1/y)>=4 suy ra 1/z=1/x+1/y>=4/x+y(do x,y>0)hay 4/4z>=4/x+y suy ra x+y>=4z.

Sau đó dùng BĐT Bunhiacopxki suy ra 2(√x+√y)^2>=(x+y)^2=16z^2 suy ra

√x+√y>=√8z=2z√2

Giả sử cả 3 bđt trên đều đúng, như vậy  \(a\left(1-a\right).b\left(1-b\right).c\left(1-c\right)>\frac{1}{4}.\frac{1}{4}.\frac{1}{4}=\frac{1}{64}\)

Mặt khác vì  \(0< a,b,c< 1\)  nên:

\(0< a\left(1-a\right)=-a^2+a-\frac{1}{4}+\frac{1}{4}=\frac{1}{4}-\left(a-\frac{1}{2}\right)^2\le\frac{1}{4}\)

Tương tự  \(0< b\left(1-b\right)\le\frac{1}{4}\)  và  \(0< c\left(1-c\right)\le\frac{1}{4}\)

Suy ra  \(a\left(1-a\right).b\left(1-b\right).c\left(1-c\right)\le\frac{1}{4}.\frac{1}{4}.\frac{1}{4}=\frac{1}{64}\)  (vô lý)

Vậy phải có ít nhất 1 bđt sai

Ta có \(\hept{\begin{cases}\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\forall x;y>0\\\left(\sqrt{y}-\sqrt{z}\right)^2\ge0\forall y;z>0\\\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\forall z;x>0\end{cases}}\)

\(\Rightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\)

\(\Leftrightarrow x-2\sqrt{xy}+y+y-2\sqrt{yz}+z+z-2\sqrt{zx}+x\ge0\)

\(\Leftrightarrow2\left(x+y+z\right)-2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\ge0\)

\(\Leftrightarrow x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

Vậy .....