Lời giải:

$3x^2-4xy+y^2=0$

$\Leftrightarrow 3x(x-y)-y(x-y)=0$

$\Leftrightarrow (x-y)(3x-y)=0$
$\Rightarrow x-y=0$ hoặc $3x-y=0$

Nếu $x-y=0\Leftrightarrow x=y$. Thay vào pt $(2)$:
$x^2+2x=8$

$\Leftrightarrow x^2+2x-8=0$

$\Leftrightarrow (x-2)(x+4)=0$

$\Rightarrow x=2$ hoặc $x=-4$. 

Vậy hpt có nghiệm $(x,y)=(2,2); (-4,-4)$

Nếu $3x-y=0$

$\Leftrightarrow 3x=y$. Thay vô pt $(2)$:

$x^2+6x=8$

$\Leftrightarrow x^2+6x-8=0$
$\Rightarrow x=-3\pm \sqrt{17}$

$\Rightarrow y=3(-3\pm \sqrt{17})$ (tương ứng) 

Vậy tổng cộng hpt có 4 nghiệm $(x,y)$ thực.

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

Lời giải:

Đặt \((\sqrt{x-y},\sqrt{x+y})=(b,a)\)

HPT trở thành: \(\left\{\begin{matrix} a-b=2(1)\\ \sqrt{\frac{a^4+b^4}{2}}+ab=4(2)\end{matrix}\right.\)

\((2)\Leftrightarrow \sqrt{\frac{a^4+b^4}{2}}=4-ab\). Bình phương hai vế:
\(\Rightarrow \frac{a^4+b^4}{2}=16+a^2b^2-8ab\)

\(\Leftrightarrow a^4+b^4-2a^2b^2=32-16ab\)

\(\Leftrightarrow (a^2-b^2)^2=32-16ab\Leftrightarrow 4(a+b)^2=32-16ab\) (do \(a-b=2\) )

\(\Leftrightarrow (a+b)^2=8-4ab\)

Thay \(a=b+2\Rightarrow (2b+2)^2=8-4b(b+2)\)

\(\Leftrightarrow (b+1)^2=2-b(b+2)\Leftrightarrow 2b^2+4b-1=0\)

\(\Rightarrow b=\frac{-2+\sqrt{6}}{2}\) (do \(b\geq 0\))

Từ đó kéo theo \(a=\frac{2+\sqrt{6}}{2}\). Từ đây suy ra \((x,y)=(\frac{5}{2},\sqrt{6})\)

1)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

x= 13/7  

y=5/7

`{(2x-y=3),(x+3y=4):}`

`<=>{(6x-3y=9),(x+3y=4):}`

`<=>{(7x=13),(x+3y=4):}`

`<=>{(x=13/7),(13/7+3y=4):}`

`<=>{(x=13/7),(y=5/7):}`

2x - y = 3

x + 3y = 4

6x - 3y = 9

x + 3y = 4

7x = 13

x + 3y = 4

x = 13/7

y = 5/7