a)

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

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

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

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

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

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

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

Vậy hệ phương trình có tập nghiệm {1;-2}

b)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}x-1\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}.\left(\frac{3\sqrt{3}+2\sqrt{2}}{19}\right)-1\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-10+2\sqrt{6}}{19}\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\\y=\frac{-10+2\sqrt{6}}{19}\end{matrix}\right.\)

Vậy hệ phương trình có tập nghiệm \(\left\{\frac{3\sqrt{3}+2\sqrt{2}}{19};\frac{-10+2\sqrt{6}}{19}\right\}\)

c)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}7x=13\\4x+2y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{13}{7}\\4.\frac{13}{7}+2y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{13}{7}\\y=\frac{9}{7}\end{matrix}\right.\)

Vậy hệ phương trình có tập nghiệm \(\left\{\frac{13}{7};\frac{9}{7}\right\}\)

Lời giải:

a)

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

Lấy $(1)$ trừ $(2)$:

$\Rightarrow 2x=-1\Rightarrow x=-\frac{1}{2}$

Thay $x=\frac{-1}{2}$ vào $(1):y=5x-4=5.\frac{-1}{2}-4=\frac{-13}{2}$

Vậy HPT có nghiệm $(x,y)=(\frac{-1}{2}, \frac{-13}{2})$

b)

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

Lấy $(2)-(1)$ thu được:

$5y=3-\sqrt{2}\Rightarrow y=\frac{3-\sqrt{2}}{5}$

Thay giá trị $y$ trên vào $(1): x=\frac{2y+\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{6}+\sqrt{3}}{5}$

Vậy.........

\(\left\{{}\begin{matrix}15x=y-5\\16x=y+3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}15x=y-5\\x=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=125\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{y\sqrt{2}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-24+10\sqrt{6}\\y=5-2\sqrt{6}\end{matrix}\right.\)

a. ĐK: \(x\ge1;y\ge1\)

Đặt \(\sqrt{x-1}=a\left(a\ge0\right)\) và \(\sqrt{y-1}=b\left(b\ge0\right)\)

Khí đó hệ phương trình trở thành:

\(\left\{{}\begin{matrix}2a-b=1\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2a-1\\a+2a-1=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=2.1-1\\a=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=1\end{matrix}\right.\)(tm)

* a = 1 \(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)(tmđk)

* b = 1 \(\sqrt{y-1}=1\Leftrightarrow y-1=1\Leftrightarrow y=2\) (tmđk)

Vậy nghiệm của hệ phương trình là (2;2)

b. Đặt \(\left(x-1\right)^2=a\) ( a \(\ge\) 0)

Khi đó hệ phương trình đã cho trở thành :

\(\left\{{}\begin{matrix}a-2y=2\\3a+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2+2y\\3\left(2+2y\right)+3y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2+2.\left(-\dfrac{5}{9}\right)\\y=-\dfrac{5}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{9}\\y=-\dfrac{5}{9}\end{matrix}\right.\)(tmđk)

* a = \(\dfrac{8}{9}\Leftrightarrow\) \(\left(x-1\right)^2=\dfrac{8}{9}=\left(\pm\dfrac{2\sqrt{2}}{3}\right)^2\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{2}}{3}+1\\x=-\dfrac{2\sqrt{2}}{3}+1\end{matrix}\right.\)

Vậy nghiệm của hệ phương trình là \(\left(\dfrac{2\sqrt{2}}{3};-\dfrac{5}{9}\right);\left(\dfrac{-2\sqrt{2}}{3};-\dfrac{5}{9}\right)\)

a/\(\left\{{}\begin{matrix}\sqrt{5}-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x-3\sqrt{5}=15\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)

\(\Leftrightarrow15x+2\sqrt{3}x=15\left(\sqrt{3}-1\right)+21=15\sqrt{3}+6\)

\(\Leftrightarrow x=\frac{15\sqrt{3}+6}{15+2\sqrt{3}}=\sqrt{3}\)

\(\Rightarrow y=\sqrt{5}\)

Kết luận nghiệm pt: \(\left\{{}\begin{matrix}x=\sqrt{3}\\y=\sqrt{5}\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}7x=4y\\x-y+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-4y=0\\7x-7y+21=0\end{matrix}\right.\)

\(\Leftrightarrow\left(7x-4y\right)-\left(7x-7y+21\right)=0\)

\(\Leftrightarrow3x-21=0\Leftrightarrow x=7\)

\(\Rightarrow y=4\)

Kết luận nghiệm pt: \(\left\{{}\begin{matrix}x=7\\y=4\end{matrix}\right.\)

áp dụng phương pháp thế nhé bạn.