\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^3-2y^3=30\\5\left(x-y\right)\left(x^2+2y^2\right)=30\end{matrix}\right.\)

Trừ vế cho vế:

\(5\left(x-y\right)\left(x^2+2y^2\right)-\left(4x^3-2y^3\right)=0\)

\(\Leftrightarrow x^3-5x^2y+10xy^2-8y^3=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x^2-3xy+4y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=y=0\left(ktm\right)\end{matrix}\right.\)

Thay vào pt đầu:

\(\Rightarrow2\left(2y\right)^3-y^3=15\)

\(\Rightarrow y^3=1\Rightarrow y=1\Rightarrow x=2\)

các bn ơi giúp mình với

a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}\\6x+2y=-2\end{matrix}\right.-6x+12y=18}\)

Lời giải:

HPT \(\Leftrightarrow \left\{\begin{matrix} 2(\sqrt{5}+2)x+2y=6-2\sqrt{5}\\ -x+2y=6-2\sqrt{5}\end{matrix}\right.\)

Lấy PT(1) trừ PT(2) theo vế:

$\Rightarrow 2(\sqrt{5}+2)x+x=(6-2\sqrt{5})-(6-2\sqrt{5})$

$\Leftrightarrow (2\sqrt{5}+5)x=0$

$\Leftrightarrow x=0$

$y=3-\sqrt{5}-(\sqrt{5}+2)x=3-\sqrt{5}-(\sqrt{5}+2).0=3-\sqrt{5}$

Lời giải:

Lấy 2 PT trừ theo vế thì:

$x^3-y^3=x-y$

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

$\Leftrightarrow (x-y)(x^2+xy+y^2-1)=0$

$\Rightarrow x-y=0$ hoặc $x^2+xy+y^2=1$
TH1: $x-y=0\Leftrightarrow x=y$

Thay vào PT(1):

$x^3=3x\Leftrightarrow x(x^2-3)=0\Leftrightarrow x=0$ hoặc $x=\pm \sqrt{3}$

Vậy $(x,y)=(0,0), (\sqrt{3}, \sqrt{3}), (-\sqrt{3}, -\sqrt{3})$

TH2: $x^2+xy+y^2=1(*)$

Cộng 2 PT theo vế: $x^3+y^3=3(x+y)$

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

Nếu $x+y=0$ thì $x=-y$. Thay vào $(*)$:

$x^2+x(-x)+y^2=1$

$\Leftrightarrow y^2=1\Leftrightarrow y=\pm 1$

Vậy $(x,y)=(1,-1), (-1,1)$

Nếu $x^2-xy+y^2-3=0$

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

$\Leftrightarrow 1-2xy-3=0$

$\Leftrightarrow xy=-1$

$x^2+y^2=1-xy=1-(-1)=2$

$\Leftrightarrow (x+y)^2-2xy=2$

$\Leftrightarrow (x+y)^2-2(-1)=2$

$\Leftrightarrow x+y=0$

$\Leftrightarrow x=-y$. Thay vào $xy=-1$ thì: $y^2=1\Leftrightarrow y=\pm 1$

Nếu $y=1$ thì $x=-y=-1$. Nếu $y=-1$ thì $x=-y=1$

Vậy $(x,y)=(-1,1), (1,-1)$.

Vậy............

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

⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))