Lời giải:

ĐKXĐ:.......

$PT\Leftrightarrow \frac{4}{x}-x=\sqrt{2x-\frac{5}{x}}-\sqrt{x-\frac{1}{x}}$

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

$\Leftrightarrow \frac{4}{x}-x = \frac{x-\frac{4}{x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}$

$\Leftrightarrow (\frac{4}{x}-x)\left[1+\frac{1}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}\right]=0$

Hiển nhiên biểu thức trong ngoặc vuông luôn dương nên $\frac{4}{x}-x=0$

$\Rightarrow 4-x^2=0$

$\Leftrightarrow x=\pm 2$

Thử lại thấy $x=2$ thỏa mãn. 

Vậy.......

\(\Leftrightarrow x-\dfrac{4}{x}=\sqrt{x-\dfrac{1}{x}}-\sqrt{2x-\dfrac{5}{x}}\)

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

x-4/x>0

=>4/x-x<0

=>Loại

x-4/x<0

=>4/x-x>0

=>Mâu thuẫn

=>Loại

Do đó, chỉ có 1 trường hợp là x-4/x=0

=>x=2

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

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

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

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......

\(ĐKXĐ:x\ne0,x-\dfrac{1}{x}\ge0\)

Chia cả hai vế của phương trình đầu cho \(x\ne0\) ta có :

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

\(\Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)

Đặt \(\sqrt{x-\dfrac{1}{x}}=a\left(a\ge0\right)\)

Khi đó pt có dạng : \(a^2+2a-3=0\Leftrightarrow\left(a+3\right)\left(a-1\right)=0\)

\(\Leftrightarrow a=1\) ( do \(a\ge0\) )

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

\(\Leftrightarrow x=\dfrac{1\pm\sqrt{5}}{2}\) ( thỏa mãn ĐKXĐ )

\(1+\dfrac{1}{\sqrt{x^2-1}}=\dfrac{35}{12x}\left(x< -1;1< x\right)\)

Với \(x< -1\) thì pt vô nghiệm

Xét \(x>1\)

\(PT\Leftrightarrow x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}\left(nhân.x.2.vế\right)\\ \Leftrightarrow x^2+\dfrac{x^2}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}=\dfrac{1225}{144}\\ \Leftrightarrow\dfrac{x^4}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}=\dfrac{1225}{144}\\ \Leftrightarrow\left(\dfrac{x^2}{\sqrt{x^2-1}}\right)^2+\dfrac{2x^2}{\sqrt{x^2-1}}-\dfrac{1225}{144}=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2}{\sqrt{x^2-1}}=\dfrac{25}{12}\left(tm\right)\\\dfrac{x^2}{\sqrt{x^2-1}}=-\dfrac{49}{12}\left(ktm\right)\end{matrix}\right.\Leftrightarrow\dfrac{x^4}{x^2-1}=\dfrac{625}{144}\\ \Leftrightarrow144x^4-625x^2+625=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=\dfrac{5}{4}\left(tm\right)\\x=-\dfrac{5}{4}\left(tm\right)\\x=-\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

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 \(x>0;x\ne1\)

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

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

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

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

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

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

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