a) \(\left\{{}\begin{matrix}2x^2-5xy-y^2=1\\y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=1\end{matrix}\right.\)

ĐKXĐ:...

\(\Rightarrow y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=2x^2-5xy-y^2\)

Từ giả thiết dễ thấy \(y\ne0\), chia cả 2 vế cho \(y^2\) ta được:

\(\dfrac{\sqrt{xy-2y^2}+\sqrt{4y^2-xy}}{y}=\dfrac{2x^2-5xy-y^2}{y^2}\)

\(\Leftrightarrow\sqrt{\dfrac{xy-2y^2}{y^2}}+\sqrt{\dfrac{4y^2-xy}{y^2}}=2\left(\dfrac{x}{y}\right)^2-\dfrac{5x}{y}-1\)

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

Đặt \(\dfrac{x}{y}=t\) \(\left(2\le t\le4\right)\)

\(\Leftrightarrow\sqrt{t-2}+\sqrt{4-t}=2t^2-5t-1\)

\(\Leftrightarrow\sqrt{t-2}-1+\sqrt{4-t}-1=2t^2-5t-3\)

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

\(\Leftrightarrow\left(t-3\right)\left(2t+1-\dfrac{1}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}\right)=0\)

Xét \(2t+1-\dfrac{1}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}=2t+\dfrac{\sqrt{t-2}}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}>0\forall t\)

\(\Rightarrow t-3=0\)

\(\Leftrightarrow t=3\)

\(\Leftrightarrow\dfrac{x}{y}=3\Leftrightarrow x=3y\)

Thế vào phương trình \(\left(1\right):2\cdot9y^2-5y\cdot3y-y^2-1=0\)

\(\Leftrightarrow2y^2-1=0\)

\(\Leftrightarrow y=\dfrac{1}{\sqrt{2}}\) do \(y>0\)

\(\Leftrightarrow x=\dfrac{3}{\sqrt{2}}\)

Vậy tập nghiệm của phương trình \(\left(x;y\right)=\left(\dfrac{3}{\sqrt{2}};\dfrac{1}{\sqrt{2}}\right)\)

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

Trừ theo vế 2 phương trình ta được:

\(x^3-y^3=2\left(x^2-y^2-2x+2y\right)\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-2\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-2\left(x+y\right)+4\right)=0\)

Xét phương trình \(x^2+x\left(y-2\right)+y^2-2y+4=0\)

\(\Delta_x=\left(y-2\right)^2-4\left(y^2-2y+4\right)=-3y^2+4y-8< 0\) nên phương trình vô nghiệm.

Do đó \(x=y\)

Thế vào phương trình \(\left(1\right):x^3+1=2x^2\)

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

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

Vậy...

\(\begin{cases}\left(x+y+3\right)\cdot\sqrt{x+2y}+2y+4=0\left(1\right)\\ \left(x-y\right)\left(x^2+4\right)=y^2+1\left(2\right)\end{cases}\)

ĐKXĐ: x-2y>=0

Đặt \(t=\sqrt{x-2y}\) (Điều kiện: t>=0)

=>\(x-2y=t^2\)

=>\(x=2y+t^2\)

Thay \(x=2y+t^2\) vào (1), ta được:

\((t^2 + 2y + y + 3)t + 2y + 4 = 0\)

=>\((t^2 + 3y + 3)t + 2y + 4 = 0\)

=>\(y(3t + 2) = -t^3 - 3t - 4 \implies y = -\frac{t^3 + 3t + 4}{3t + 2}\)

=>\(x = t^2 + 2\left(-\frac{t^3 + 3t + 4}{3t + 2}\right) = \frac{t^3 + 2t^2 - 6t - 8}{3t + 2}\)

\(x - y = \frac{t^3 + 2t^2 - 6t - 8 + t^3 + 3t + 4}{3t + 2} = \frac{2t^3 + 2t^2 - 3t - 4}{3t + 2}\)

Thay x và y vào (2), ta được:

\((2t^3 + 2t^2 - 3t - 4)[(t^3 + 2t^2 - 6t - 8)^2 + 4(3t + 2)^2] = (3t + 2)[(t^3 + 3t + 4)^2 + (3t + 2)^2]\)

=>\(t = \sqrt{3}\)

=>\(y = -\frac{(\sqrt{3})^3 + 3\sqrt{3} + 4}{3\sqrt{3} + 2} = -\frac{6\sqrt{3} + 4}{3\sqrt{3} + 2} = -2\) và \(x = (\sqrt{3})^2 + 2(-2) = 3 - 4 = -1\)

- Trừ hai pt ta được :\(x^3-y^3-x^2+y^2+x-y+1-1=2y-2x\)

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

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

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

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

TH1 : x = y

PT ( I ) TT : \(x^3-x^2+x+1-2x=x^3-x^2-x+1=0\)

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

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

\(\Leftrightarrow x=y=\pm1\)

TH₂ : \(x^2+xy+y^2-x-y+3=0\)

\(\Leftrightarrow x^2+\dfrac{y^2}{4}+\dfrac{1}{4}+xy-x-\dfrac{1}{2}y+\dfrac{3}{4}y^2-\dfrac{1}{2}y+\dfrac{11}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2+\dfrac{8}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2=-\dfrac{8}{3}\left(VL\right)\)

Vậy ....

6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)

7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)

8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y

(Các câu khác tương tự nhé.)

b: \(\begin{cases}81x^3y^2-81x^2y^2+33xy^2-29y^2=4\left(1\right)\\ 25y^3+9x^2y^3-6xy^3-4y^2=24\left(2\right)\end{cases}\)

Khi y=0 thì (2) sẽ trở thành: \(25\cdot0^3+9x^2\cdot0^3-6\cdot x\cdot0^3-4\cdot0^2=24\)

=>0=24(loại)

=>y<>0

Ta có: \(81x^3y^2-81x^2y^2+33xy^2-29y^2=4\)

=>\(y^2(81x^3-81x^2+33x-29)=4\)

=>\(81x^3-81x^2+33x-29=\frac{4}{y^2}\)

Ta có: \(25y^3+9x^2y^3-6xy^3-4y^2=24\)

=>\(y^3(9x^2 - 6x + 25) - 4y^2 = 24\)

=>\(y(9x^2 - 6x + 25) - 4 = \frac{24}{y^2}\)

Đặt u=3x-1

=>3x=u+1

=>\(x=\frac{u+1}{3}\)

\(9x^2-6x+25=9x^2-6x+1+24=\left(3x-1\right)^2+24\)

\(=u^2+24\)

(1) sẽ tương đương: \(81\left(\frac{u+1}{3}\right)^3 - 81\left(\frac{u+1}{3}\right)^2 + 33\left(\frac{u+1}{3}\right) - 29 = \frac{4}{y^2}\)

=>\(3(u+1)^3 - 9(u+1)^2 + 11(u+1) - 29 = \frac{4}{y^2}\)

=>\(3u^3+2u-24=\frac{4}{y^2}\quad(3)\)

THay (3) vào phương trình (2), ta được:

\(y(u^2+24)-4=\frac{24}{y^2}\quad(4)\)

Ta có: \(3u^3+2u-24=\frac{4}{y^2}\quad(3)\)

=>\(18u^3+12u-144=\frac{24}{y^2}\)

=>\(y(u^2 + 24) - 4 = 18u^3 + 12u - 144\)

=>\(y(u^2+24)=18u^3+12u-140\)

=>\(y=\frac{18u^3 + 12u - 140}{u^2 + 24}\)

Thay lại vào (3), ta được: \(\frac{4}{\left(\frac{18u^3 + 12u - 140}{u^2 + 24}\right)^2} = 3u^3 + 2u - 24\)

=>\((9u^3 + 6u - 70)^2(3u^3 + 2u - 24) = (u^2 + 24)^2\)

=>u=2

\(y=\frac{18\cdot2^3+12\cdot2-140}{2^2+24}=\frac{144 + 24 - 140}{28}=\frac{28}{28}=1\)

3x-1=2

=>3x=3

=>x=1

b: \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y^2+18y+9+y^2-6y-6-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10y^2+10y-20=0\\x=3y+3\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}y\in\left\{-2;1\right\}\\x=3y+3\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(-3;-2\right);\left(6;1\right)\right\}\)

a: \(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y=4x-6\\3x^2+6xy-x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{9}x-\dfrac{2}{3}\\3x^2+6x\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)-x+3\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x^2+\dfrac{8}{3}x^2-4x-x+\dfrac{4}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{17}{3}x^2-\dfrac{11}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x^2-11x-6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(17x+6\right)=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}17x+6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\cdot1-\dfrac{2}{3}=\dfrac{4}{9}-\dfrac{2}{3}=-\dfrac{2}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{6}{17}\\y=\dfrac{4}{9}\cdot\dfrac{-6}{17}-\dfrac{2}{3}=\dfrac{-14}{17}\end{matrix}\right.\end{matrix}\right.\)