Giải hệ pt sau \(\left\{{}\begin{matrix}x^2-xy+y^2=3\\z^2+yz+1=0\end{matrix}\right.\)

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

Đặt \(\sqrt{2x+1}=t\ge0\Rightarrow x=\dfrac{t^2-1}{2}\)

\(\dfrac{5\left(t^2-1\right)}{2}-\left(\dfrac{t^2-1}{2}+4\right)t-4=0\)

\(\Leftrightarrow t^3-5t^2+7t+13=0\)

\(\Leftrightarrow\left(t+1\right)\left(t^2-6t+13\right)=0\)

Còn nếu sửa đề thành: \(5x-\left(x+4\right)\sqrt{2x+1}+4=0\)

Thì sau khi đặt ẩn phụ như trên, pt trở thành: \(\left(t-3\right)\left(t-1\right)^2=0\)

a: =>3x^2-6x-x+2=0

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

b: =>x^4-x-4x+4=0

=>x(x-1)(x^2+x+1)-4(x-1)=0

=>(x-1)(x^3+x^2+x-4)=0

=>x-1=0 hoặc x^3+x^2+x-4=0

=>x=1 hoặc x=1,15

\(m=-4\Leftrightarrow x^4+5x^2-6=0\\ \Leftrightarrow x^4+6x^2-x^2-6=0\\ \Leftrightarrow\left(x^2+6\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)