c.

\(\left\{\begin{matrix} 9x-6y=4\\ 15x-10y=7\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{6y+4}{9}\\ 15x-10y=7\end{matrix}\right.\)

\(\Rightarrow 15.\frac{6y+4}{9}-10y=7\)

\(\Leftrightarrow \frac{5}{3}(6y+4)-10y=7\Leftrightarrow \frac{20}{3}=7\) (vô lý)

Do đó hpt vô nghiệm.

d.

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

\(\Leftrightarrow 17y+20=3\Leftrightarrow 17y=-17\Leftrightarrow y=-1\)

\(x=3y+5=-3+5=2\)

Vậy HPT có nghiệm $(x,y)=(2,-1)$

Các câu còn lại bạn làm theo pp tương tự.

1.

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

\(\Rightarrow 5x-(3x-5)=4\Leftrightarrow 2x+5=4\Leftrightarrow 2x=-1\Leftrightarrow x=\frac{-1}{2}\)

\(y=3x-5=\frac{-3}{2}-5=\frac{-13}{2}\)

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

2.

HPT \(\Leftrightarrow \left\{\begin{matrix} 4xy+8x-6y-12=4xy-12x+54\\ 3xy-3x+3y-3=3xy+3y-12\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} 20x-6y=66\\ -3x=-9\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 6y=20x-66\\ x=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3\\ y=-1\end{matrix}\right.\)

Vậy HPT có nghiệm $(3,-1)$

3.

HPT\(\Leftrightarrow \left\{\begin{matrix} (x+2)(y+3)-xy=100\\ xy-(x-2)(y-2)=64\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3x+2y=94\\ 2x+2y=68\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} 3x+2y=94\\ 2y=68-2x\end{matrix}\right.\Rightarrow 3x+68-2x=94\)

\(\Leftrightarrow x+68=94\Leftrightarrow x=26\)

\(y=\frac{68-2x}{2}=8\)

Vậy hpt có nghiệm $(x,y)=(26,8)$

4.

HPT \(\Leftrightarrow \left\{\begin{matrix} xy-x+20y-20=xy\\ xy+x-10y-10=xy\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -x+20y=20\\ x-10y=10\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} -x+20y=20\\ x=10y+10\end{matrix}\right.\Rightarrow -10y-10+20y=20\)

\(\Leftrightarrow 10y-10=20\Leftrightarrow y=3\)

$x=10y+10=30+10=40$

Vậy hpt có nghiệm $(x,y)=(40,3)$

7.

Đặt $x^2-2x=a; \sqrt{y+1}=b$ thì hpt trở thành:
\(\left\{\begin{matrix} 2a+b=0\\ 3a-2b=-7\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} b=-2a\\ 3a-3b=-7\end{matrix}\right.\Rightarrow 3a-3(-2a)=-7\)

\(\Leftrightarrow 9a=-7\Leftrightarrow a=\frac{-7}{9}\)

\(b=-2a=\frac{14}{9}\)

Vậy: \(\left\{\begin{matrix} a=x^2-2x=\frac{-7}{9}\\ b=\sqrt{y+1}=\frac{14}{9}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2-2x+\frac{7}{9}=0\\ y+1=\frac{196}{81}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{3\pm \sqrt{2}}{3}\\ y=\frac{115}{81}\end{matrix}\right.\)

8. Đặt $|x-1|=a$ và $|y+2|=b$ thì hpt trở thành:
\(\left\{\begin{matrix} 5|x-1|-3|y+2|=7\\ 4\sqrt{x^2-2x+1}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} 5|x-1|-3|y+2|=7\\ 4\sqrt{(x-1)^2}+5\sqrt{(y+2)^2}=13\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 5|x-1|-3|y+2|=7\\ 4|x-1|+5|y+2|=13\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} 5a-3b=7\\ 4a+5b=13\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=\frac{3b+7}{5}\\ 4a+5b=13\end{matrix}\right.\)

\(\Rightarrow 4.\frac{3b+7}{5}+5b=13\Rightarrow b=1\)

$a=\frac{3b+7}{5}=\frac{10}{5}=2$

Vậy: \(\left\{\begin{matrix} a=|x-1|=2\\ b=|y+2|=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-1=\pm 2\\ y+2=\pm 1\end{matrix}\right.\)

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

Vậy $(x,y)=(3,-1), (3,-3); (-1,-1); (-1,-3)$

9. Đặt $\frac{1}{x-1}=a; \sqrt{y-2}=b$ thì hpt trở thành:

\(\left\{\begin{matrix} 3a+2b=7\\ a-b=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=1\\ b=2\end{matrix}\right.\) (sử dụng pp thế như những bài trước)

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

Vậy hpt có nghiệm $(x,y)=(2,6)$

10. Đặt $\frac{1}{\sqrt{x-1}}=a; \frac{1}{\sqrt{y+1}}=b$ thì hpt trở thành:
\(\left\{\begin{matrix} 2a+3b=5\\ 3a-2b=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=1\\ b=1\end{matrix}\right.\) (sử dụng pp thế như những bài trước)

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

Vậy hpt có nghiệm $(x,y)=(2,0)$

11.

Đặt $\sqrt{3-x}=a; \frac{1}{\sqrt{y-2}}=b$ thì hpt trở thành:
\(\left\{\begin{matrix} a+2b=5\\ 4a+3b=15\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=3\\ b=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \sqrt{3-x}=3\\ \frac{1}{\sqrt{y-2}}=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-6\\ y=3\end{matrix}\right.\)

Vậy hpt có nghiệm $(x,y)=(-6,3)$

12. Đặt $x+y=a; \sqrt{x+1}=b$ thì hpt trở thành:
\(\left\{\begin{matrix} 2a+b=4\\ a-3b=-5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=1\\ b=2\end{matrix}\right.\) (sử dụng pp thế như những bài trước)

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

Vậy hpt có nghiệm $(x,y)=(3,-2)$