a<b => |a-b|=b-a 

=>\(\frac{1}{a-b}.\sqrt{a^4\left(a-b\right)^2}=\frac{1}{a-b}.\sqrt{\left[a^2\left(a-b\right)\right]^2}=\frac{1}{a-b}.\left|a^2\left(a-b\right)\right|=\frac{1}{a-b}.a^2.\left|a-b\right|\)

\(=\frac{1}{a-b}.a^2.\left(b-a\right)=-a^2\)

A)  \(\sqrt{4\left(1-x\right)^2}-6=0\)

\(\sqrt{4\left(1-x\right)^2}=6\)

\(\hept{\begin{cases}4\left(1-x\right)=6\\4\left(1-x\right)=-6\end{cases}}\)

\(\hept{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)

B)\(\sqrt{1-12x+36x^2}=5\)

\(\sqrt{\left(1-6x\right)^2}=5\)

\(\hept{\begin{cases}1-6x=5\\1-6x=-5\end{cases}}\)

\(\hept{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

\(\frac{\sqrt{a}-2a}{2\sqrt{a}-1}=\frac{\sqrt{a}\left(1-2\sqrt{a}\right)}{2\sqrt{a}-1}=-\sqrt{a}\)

áp dụng bđt cauchy:

\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{x}.\frac{1}{y}}=\frac{2}{\sqrt{xy}}.\)

Tượng tự \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}},\frac{1}{z}+\frac{1}{x}\ge\frac{2}{\sqrt{xz}}.\)

=>2VT>=2Vp

<=>VT>=VP

dấu = xảy ra khi x=y=z

By AM-GM we have:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}};\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}};\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{xz}}\)

Cộng theo vế rồi rút gọn là có ĐPCM

Xảy ra khi x=y=z

\(5-\sqrt{x^2-6x+14}=5-\sqrt{x^2-6x+9+5}=5-\sqrt{\left(x-3\right)^2+5}\le5-\sqrt{5}\)

Dấu "=" xảy ra khi x=3