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

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

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}7x=14\\y=5-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy nghiệm duy nhất của hpt là: (2;1)

c) \(\left\{{}\begin{matrix}2y-x=2\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\2\left(2y-2\right)-y=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\4y-4-y=-1\end{matrix}\right.\)

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

Vậy nghiệm duy nhất của hpt là: (0;1)

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

Từ (1): \(x=2-2y\) (3)

Thế (3) vào (2), ta được: \(-2\left(2-2y\right)+y=1< =>-4+4y+y=1\)

                                          \(\Leftrightarrow y=1\)\(\Rightarrow\)\(x=2-2.1=0\)

Vậy nghiệm duy nhất của hpt là:  (0;1)

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

\(x^2y+2y+x=4xy< =>xy\left(x+3\right)=4xy< =>x+3=4< =>x=1\)

Thế x=1 vào 1 trong 2 phương trình => y=1

Làm mẫu hai câu a, b thôi nha.

a, \(\left\{{}\begin{matrix}x-\sqrt{3}y=0\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}y\\5y=1+\sqrt{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\approx0,95\\y\approx0,55\end{matrix}\right.\)

b, \(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{5}y=1\\x+\sqrt{5}y=\sqrt{2}\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y\approx0,19\\x=1\end{matrix}\right.\)

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

Lấy phương trình dưới trừ phương trình trên thu được: \(5y=1+\sqrt{3}\Rightarrow y=\dfrac{1+\sqrt{3}}{5}\Rightarrow x=\sqrt{3}y=\dfrac{3+\sqrt{3}}{5}\)

b) Cộng hai phương trình lại với nhau thu được:

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

c) \(\left\{{}\begin{matrix}\sqrt{2}x+\sqrt{5}y=2\\x+\sqrt{5}y=2\end{matrix}\right.\)

Lấy phương trình trên trừ phương trình dưới:

\(\left(\sqrt{2}-1\right)x=0\Leftrightarrow x=0\Rightarrow y=\dfrac{2-x}{\sqrt{5}}=\dfrac{2}{\sqrt{5}}\)

d) Hướng dẫn. Nhân phương trình đầu với \(\sqrt{2}\) rồi lấy phương trình thu được trừ phương trình dưới.

Sorry bạn, mình không nhìn kỹ đề nên nhầm phương pháp.