Chắc là giải hệ phương trình?

a.

\(\left\{{}\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)

Xét pt: \(x^2+x-xy-2y^2-2y=0\)

\(\Leftrightarrow\left(x^2-xy-2y^2\right)+x-2y=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)+\left(x-2y\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x-2y\right)=0\)

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

TH1: \(x=-y-1\) thế vào \(x^2+y^2=1\)

\(\Rightarrow\left(-y-1\right)^2+y^2=1\)

\(\Leftrightarrow2y^2+2y=0\Rightarrow\left[{}\begin{matrix}y=0\Rightarrow x=-1\\y=-1\Rightarrow x=0\end{matrix}\right.\)

TH2: \(x=2y\) thế vào \(x^2+y^2=1\)

\(\Rightarrow\left(2y\right)^2+y^2=1\Leftrightarrow5y^2=1\)

\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{1}{\sqrt{5}}\Rightarrow x=\dfrac{2}{\sqrt{5}}\\y=-\dfrac{1}{\sqrt{5}}\Rightarrow x=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}6x^2-3xy+x=1-y\\x^2+y^2=1\end{matrix}\right.\)

Xét pt: \(6x^2-3xy+x=1-y\)

\(\Leftrightarrow\left(6x^2+x-1\right)-3xy+y=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x+1\right)-y\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x+1-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\y=2x+1\end{matrix}\right.\)

Thế vào \(x^2+y^2=1\) tương tự câu a...

sau bạn đăng tách ra cho mn cùng giúp nhé 

a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)

b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)

c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)

d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)

e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)

f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)