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

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

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

Biểu thức <=> \(\left(-\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

\(H=\frac{4}{\sqrt{x}+1}+\frac{2}{1-\sqrt{x}}-\frac{\sqrt{x}-5}{x-1}\)

\(=\frac{4}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{4\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(4\sqrt{x}-4\right)-\left(2\sqrt{x}+2\right)-\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(H=\frac{4}{\sqrt{x}+1}+\frac{2}{1-\sqrt{x}}-\frac{\sqrt{x}-5}{x-1}\)

\(=\frac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}+1}\)

ĐK : \(\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

\(P=\frac{2x}{\sqrt{x}-2}=\frac{2x-8+8}{\sqrt{x}-2}=\frac{2\left(x-4\right)+8}{\sqrt{x}-2}\)

\(=2\left(\sqrt{x}+2\right)+\frac{8}{\sqrt{x}-2}=2\left(\sqrt{x}-2\right)+\frac{8}{\sqrt{x}-2}+8\)

Áp dụng BĐT Cauchy cho các số dương

\(2\left(\sqrt{x}-2\right)+\frac{8}{\sqrt{x}-2}\ge2\sqrt{2\left(\sqrt{x}-2\right).\frac{8}{\sqrt{x}-2}}=2\sqrt{16}=8\)

\(\Leftrightarrow P\ge16\). Đẳng thức xảy ra <=> x = 16

Vậy ...

\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{\left(x+4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}\)

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

\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\) (*) vì x>=4

Với\(\sqrt{x-4}-2\ge0\Rightarrow\sqrt{x-4}\ge2\Leftrightarrow x\ge8\)

Khi đó (*) \(=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

Với \(\sqrt{x-4}-2< 0\Rightarrow\sqrt{x-4}< 2\)mà x>=4 \(\Rightarrow4\le x< 8\)

Khi đó (*) \(=\sqrt{x-4}+2-\sqrt{x-4}+2=4\)

Vậy...

Chú ý: Dự đoán \(MaxP=\frac{1}{4}\)khi a = b = c = 1. Ta sẽ chứng minh đây là giá trị lớn nhất của P

Áp dụng bất đẳng thức Cauchy-Schwarz, ta được: \(\sqrt{\left(1+1+1+1\right)\left(a^2+b^2+c^2+1\right)}\ge a+b+c+1\)\(\Rightarrow\frac{1}{\sqrt{\left(a^2+b^2+c^2+1\right)}}\le\frac{2}{a+b+c+1}\)

Theo bất đẳng thức quen thuộc \(kzr\le\frac{\left(k+z+r\right)^3}{27}\), ta có: \(\frac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{2}{\frac{\left(a+b+c+3\right)^3}{27}}=\frac{54}{\left(a+b+c+3\right)^3}\)

Do đó \(P\le\frac{2}{a+b+c+1}-\frac{54}{\left(a+b+c+3\right)^3}\)

Đặt \(a+b+c=t>0\)thì ta cần chứng minh \(\frac{2}{t+1}-\frac{54}{\left(t+3\right)^3}\le\frac{1}{4}\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{\left(t-3\right)^2\left(t^2+8t+3\right)}{4\left(t+1\right)\left(t+3\right)^3}\ge0\)*đúng với mọi t > 0*

Đẳng thức xảy ra khi t = 3 hay a = b = c = 1

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

Áp dúng bất đẳng thức Cauchy-Schwarz và AM-GM, ta có

\(P\ge x+y+z+\frac{9}{x+y+z}+\frac{1}{x+y+z}\ge2\sqrt{\left(x+y+z\right).\frac{9}{x+y+z}}+\frac{1}{\sqrt{3\left(x^2+y^2+z^2\right)}}=\frac{19}{3}\)

Dấu "=" khi \(x=y=z=1\)

bình phương 2 vế ta được : 

\(x^4-4x^3+4x^2=8x-4\)

\(\Leftrightarrow x^4-4x^3+4x^2-8x+4=0\)

\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)

\(\Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)

bình phương 2 vế ta được : 

\(x^2+12-2\sqrt{x^2+12}\sqrt{x^2+5}+x^2+5=9x^2-30x+25\)

\(\Leftrightarrow2\sqrt{\left(x^2+12\right)\left(x^2+5\right)}=-7x^2+30x-8\)

\(\Leftrightarrow-45x^4-944x^2+176+420x^3+480x=0\)

\(\Leftrightarrow\left(x-2\right)\left(-45x^3+330x^2-284x-88\ne0\right)=0\)

Vậy \(x=2\)

:V bạn bình phương tới bến luôn à?