a, bình phương 2 vế ta được : 

\(-x^2+4x-3=4x^2+20x+25\)

\(\Leftrightarrow-5x^2-16x-28=0\)

\(\Delta=\left(-16\right)^2-4.\left(-28\right).\left(-5\right)=-304\)

\(x_1=\frac{-4-\sqrt{-304}}{-10};x_2=\frac{-4+\sqrt{-304}}{-10}\)

b, Ta có : \(\sqrt{3x^2-9x+1}=\left|x-2\right|\)

hay \(\sqrt{3x^2-9x+1}=\pm\left(x-2\right)\)

TH1 : \(\sqrt{3x^2-9x+1}=x-2\)

bình phương 2 vế ta được : \(3x^2-9x+1=x^2-4x+4\)

\(\Leftrightarrow2x^2-5x-3=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow x=-\frac{1}{2};3\)

TH2 : \(\sqrt{3x^2-9x+1}=-\left(x-2\right)\)

Làm tương tự 

c, \(\left(x-3\right)\sqrt{x^2+4}=x^2-9\)

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

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

\(\Leftrightarrow\left(x-3\right)\left[x^2+4-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+4-x^2-6x-9\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-5-6x\right)=0\Leftrightarrow x=3;-\frac{5}{6}\)

ĐK: \(x\ne0;y\ne0\)

\(\hept{\begin{cases}\left(x-\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=2\left(1\right)\\2x^2y+xy^2-4xy=2x-y\left(2\right)\end{cases}}\)

pt (2) \(\Leftrightarrow2x+y-4=\frac{2}{y}-\frac{1}{x}\Leftrightarrow2\left(x-\frac{1}{y}\right)+\left(y+\frac{1}{x}\right)=4\left(3\right)\)

Đặt \(\hept{\begin{cases}a=x-\frac{1}{y}\\b=y+\frac{1}{x}\end{cases}}\). Kết hớp với (1) , (3) ta có hệ \(\hept{\begin{cases}ab=2\\2a+b=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a\left(4-2a\right)=2\\b=4-2a\end{cases}\Leftrightarrow\hept{\begin{cases}a^2-2a+1=0\\b=4-2a\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\end{cases}}}}\)

Với \(\hept{\begin{cases}a=1\\b=2\end{cases}}\)ta có \(\hept{\begin{cases}x-\frac{1}{y}=1\\y+\frac{1}{x}=2\end{cases}\Leftrightarrow\hept{\begin{cases}xy-1=y\\xy+1=2x\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x-2\\x\left(2x-2\right)+1=2x\end{cases}\Leftrightarrow\hept{\begin{cases}y=2x-2\\2x^2-4x+1=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{2+\sqrt{2}}{2}\\y=\sqrt{2}\end{cases}ho\text{ặ}c\hept{\begin{cases}x=\frac{2-\sqrt{2}}{2}\\y=-\sqrt{2}\end{cases}\left(tm\right)}}\)

Vậy hệ đã cho có các nghiệm \(\left(x;y\right)=\left(\frac{2+\sqrt{2}}{2};\sqrt{2}\right);\left(\frac{2-\sqrt{2}}{2};-\sqrt{2}\right)\)

PT\(\Leftrightarrow x^3+2x=\left(2x-1\right)+2\sqrt[3]{2x-1}\)

Đặt \(\sqrt[3]{2x-1}=a\Rightarrow x^3+2x=a^3+2x\)

\(\Leftrightarrow\left(x-a\right)\left(x^2+ax+a^2+2\right)=0\Leftrightarrow\orbr{\begin{cases}\left(x^2+ax+a^2+2\right)=0\\x-a=0\end{cases}}\)

dễ thấy phương trình đầu tiên là vô nghiệm

xét \(x=a\Leftrightarrow x=\sqrt[3]{2x-1}\Leftrightarrow x^3=2x-1\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-1\pm\sqrt{5}}{2}\end{cases}}\)