Lời giải:ĐK: $\cos 3x>\frac{-1}{2}$

PT $\Rightarrow 4\sin ^2\frac{x}{2}-\sqrt{3}\cos 2x-1-2\cos ^2(x-\frac{3\pi}{4})=0$

$\Leftrightarrow 2(1-\cos x)-\sqrt{3}\cos 2x-2+[1-2\cos ^2(x-\frac{3\pi}{4})]=0$

$\Leftrightarrow -2\cos x-\sqrt{3}\cos 2x-cos (2x-\frac{3\pi}{2})=0$

$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\cos (2x-\frac{3\pi}{2})=0$

$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\sin 2x=0$

$\Leftrightarrow \cos x+\frac{\sqrt{3}}{2}\cos 2x+\frac{1}{2}\sin 2x=0$

$\Leftrightarrow \cos x-\cos (2x+\frac{5\pi}{6})=0

$\Leftrightarrow \cos x=\cos (2x+\frac{5\pi}{6})$

$\Rightarrow x+2k\pi =2x+\frac{5}{6}\pi$ hoặc $-x+2k\pi =2x+\frac{5}{6}\pi$

Vậy......

\(2sinx+2\sqrt{3}cosx-\sqrt{3}sin2x+cos2x=\sqrt{3}cosx+cos2x-2sinx+2\)

\(\Leftrightarrow4sinx+\sqrt{3}cosx-2\sqrt{3}sinx.cosx-2=0\)

\(\Leftrightarrow-2sinx\left(\sqrt{3}cosx-2\right)+\sqrt{3}cosx-2=0\)

\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{3}cosx-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=\dfrac{2}{\sqrt{3}}>1\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ:x khác 0

Xét VT=\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=8\left(x^2+\dfrac{1}{x^2}+2\right)-8\left(x^2+\dfrac{1}{x^2}\right)=16\)

=>(x+4)²=16

<=>x+4=4 hoặc x+4=-4

<=>x=0(L) hoặc x=-8(TM)

Vậy...

\(\left(x^2-3x+2\right)\sqrt{\dfrac{x+3}{x-1}}=-\dfrac{1}{2}x^3+\dfrac{15}{2}x-11\left(1\right)\)

Đk: \(\sqrt{\dfrac{x+3}{x-1}}\ge0\Leftrightarrow\left[{}\begin{matrix}x>1\\x\le-3\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow-2\left(x-1\right)\left(x-2\right)\sqrt{\dfrac{x+3}{x-1}}=x^3-15x+22\)

\(\Rightarrow-2\sqrt{\left(x-1\right)\left(x+3\right)}.\left(x-2\right)=\left(x-2\right)\left(x^2+2x-11\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\-2\sqrt{\left(x-1\right)\left(x+3\right)}=x^2+2x-11\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow-2\sqrt{x^2+2x-3}=\left(x^2+2x-3\right)-8\)

Đặt \(a=\sqrt{x^2+2x-3}\left(a\ge0\right)\). Từ phương trình (2) suy ra:

\(a^2+2a-8=0\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-4\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+2x-3}=2\Leftrightarrow x^2+2x-7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1+2\sqrt{2}\left(nhận\right)\\x=-1-2\sqrt{2}\left(nhận\right)\end{matrix}\right.\)

Thử lại ta có \(x=2\) và \(x=-1+2\sqrt{2}\) là 2 nghiệm của phương trình (1).

\(\Leftrightarrow2\left(x^2-3x+2\right)\cdot\sqrt{\dfrac{x+3}{x-1}}=-x^3+15x-22\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)\cdot\dfrac{\sqrt{\left(x+3\right)\left(x-1\right)}}{x-1}=-x^3+2x^2-2x^2+4x+11x-22\)

\(\Leftrightarrow2\left(x-2\right)\sqrt{\left(x+3\right)\left(x-1\right)}=\left(x-2\right)\left(-x^2-2x+11\right)\)

\(\Leftrightarrow\left(x-2\right)\left(\sqrt{4\left(x^2+2x-3\right)}+x^2+2x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\left(1\right)\\2\sqrt{x^2+2x-3}+x^2+2x-11=0\left(2\right)\end{matrix}\right.\)

(1) =>x=2

(2): Đặt \(\sqrt{x^2+2x-3}=a\left(a>=0\right)\)

=>2a+a^2-8=0

=>(a+4)(a-2)=0

=>a=2

=>x^2+2x-3=4

=>x^2+2x-7=0

=>\(x=-1\pm2\sqrt{2}\)

Đặt VT là T

Áp dụng AM-GM cho 3 số dương, ta có:

\(\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)

\(T\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)(đpcm)

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

\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{.....}+\dfrac{x+2}{....}\)

\(=\dfrac{\sqrt{x^3}+2x+2\sqrt{x}-2+x+2}{.....}=\dfrac{\sqrt{x^3}+3x+2\sqrt{x}}{....}\)

\(=\dfrac{\sqrt{x}\left(x+3\sqrt{x}+2\right)}{....}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{....}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

P/S: Chú ý điều kiện khi rút gọn, tự tìm.

2)

P = \(\dfrac{x}{x-\sqrt{x}}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\) với \(x>0;x\ne1\)

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

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

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

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

= \(\dfrac{x\sqrt{x}+3x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Đặt \(\sqrt{x^3-4}=a>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2=x^3-4\\a^3=\sqrt[3]{\left(x^2+4\right)^2}+4\end{matrix}\right.\)

\(\Rightarrow a^3+\sqrt[3]{\left(a^2+4\right)^2}=\sqrt[3]{\left(x^2+4\right)^2}+4+x^2\)

\(\Leftrightarrow a^3+\sqrt[3]{\left(a^2+4\right)^2}=\sqrt[3]{\left(x^2+4\right)^2}+x^3-a^2+x^2\)

\(\Leftrightarrow a^3+a^2+\sqrt[3]{\left(a^2+4\right)^2}=x^3+x^2+\sqrt[3]{\left(x^2+4\right)^2}\)

\(\Leftrightarrow a=x\)

\(\Leftrightarrow\sqrt{x^3-4}=x\)

\(\Leftrightarrow x^3-4=x^2\)

\(\Leftrightarrow x=2\)

chtt

Đặt \(\sqrt{x^3-4}=a\) để loại cai bình phương ở VP rồi biêt đổi ti thì ra. Không thich thì co thể nhân liên hiệp cũng được nhưng hơi dài.