ĐK: \(x\ge2\)

<=>\(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{\left(x-1\right)\left(x+3\right)}+\sqrt{x+3}-\sqrt{x-2}=0\)

<=> \(\sqrt{x-1}\left(\sqrt{x-2}-\sqrt{x-3}\right)-\left(\sqrt{x-2}-\sqrt{x-3}\right)=0\)

<=> \(\left(\sqrt{x-2}-\sqrt{x-3}\right)\left(\sqrt{x-1}-1\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x-2}=\sqrt{x-3}\left(loai\right)\\\sqrt{x-1}=1\end{cases}}\)

<=> x -1 = 1

<=> x = 2 ( tm đk)

Cái này đặt ẩn rồi làm như cô Chi sẽ dễ nhìn hơn.

Áp dụng BĐT AM-GM: $VP\leq \frac{25}{yz+zx+xy+4}$

Cần c/m: $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}$\leq \frac{25}{yz+zx+xy+4}$

$\Leftrightarrow (yz+zx+xy)(xy^{2}+yz^{2}+zx^{2})+4(xy^{2}+yz^{2}+zx^{2})\leq 25xyz+4(yz+zx+xy)+16$

BĐT trên sẽ được c/m nếu c/m được: $xy^{2}+yz^{2}+zx^{2}\leq 4$.

KMTTQ, g/sử y nằm giữa x và z. $\Rightarrow x(x-y)(y-z)\geq 0$

$\Leftrightarrow xy^{2}+yz^{2}+zx^{2}\leq y(x^{2}+xz+z^{2})\leq y(x+z)^{2}$

Đến đây áp dụng BĐT AM-GM:

$y(x+z)^{2}=4.y.(\frac{x+z}{2})(\frac{x+z}{2})\leq \frac{4(y+\frac{x+z}{2}+\frac{x+z}{2})^{3}}{27}=\frac{4(x+y+z)^{3}}{27}=4$ (đpcm)

Dấu bằng xảy ra khi, chẳng hạn $x=0;y=1;z=2$

Áp dụng BĐT AM-GM và BĐT Rearrangement  ta có:

\(VT=\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\)

\(=\frac{\left(x+y+z\right)^2+3\left(x+y+z\right)+xy^2+yz^2+zx^2+3}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)\(\le\frac{21+y\left(x+z\right)^2}{3\sqrt[3]{4\left(xy+yz+xz\right)}}\le\frac{21+\frac{\left(\frac{2\left(x+y+z\right)}{3}\right)^3}{2}}{3\sqrt[3]{4\left(xy+yz+zx\right)}}=\frac{21+4}{3\sqrt[3]{4\left(xy+yz+zx\right)}}=\frac{25}{3\sqrt[3]{4\left(xy+yz+zx\right)}}\)

Dấu "=" xảy ra <=> (x;y;z)=(2;1;0) và hoán vị của nó

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)

\(A=\sqrt{x}+1-\frac{17}{1-\sqrt{x}}\)

\(\Leftrightarrow A=\frac{x-1+17}{\sqrt{x}-1}\)

\(\Leftrightarrow A=\frac{x+16}{\sqrt{x}-1}\)

\(B=\frac{x-7}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-3}\)

\(\Leftrightarrow B=\frac{x-7+\sqrt{x}-3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow B=\frac{x-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow B=\frac{\sqrt{x}+3}{\sqrt{x}-1}\)

Vậy \(P=\frac{x-16}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)

\(\Leftrightarrow P=\frac{\left(x-16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{x-16}{\sqrt{x}+3}\)

b) Ta có : \(\sqrt{x}+3\ge3>0\)

Để P min \(\Leftrightarrow x-16\) min

Mà \(x-16\ge-16\)

Dấu " = " xảy ra \(\Leftrightarrow x=0\)

Vậy \(Min_P=\frac{-16}{3}\Leftrightarrow x=0\)