Đặt \(\dfrac{1}{x+1}\) = a; \(\dfrac{1}{y}\) = b (x \(\ne\) -1; y \(\ne\) 0)

Khi đó hpt trên tương đương:

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=-4\\8a+9b=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-b=1\\8a+9b=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=-1\\8a+9\left(-1\right)=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=-1\\8a=4\end{matrix}\right.\)

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

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

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

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

Vậy hpt có nghiệm duy nhất (x; y) = (1; -1)

Chúc bn học tốt!

ĐK:  ( x ≠ 1 ; y ≠ 0 ) 

Đặt a = \(\dfrac{1}{x+1} \) ; b = \(\dfrac{1}{y}\) . Ta có hệ phương trình 

\(\begin{cases} a + b = \dfrac{-1}{2}\\ 8a + 9b = -5 \end{cases} \)

⇔\(\begin{cases} 8a + 8b = -4 \\ 8a + 9b = -5 \end{cases} \) ⇔ \(\begin{cases} -b = 1 \\ a + b = \dfrac{-1}{2} \end{cases} \) ⇔ \(\begin{cases} b = - 1 \\ a = \dfrac{1}{2} \end{cases} \)

=> \(​​​​\begin{cases} \dfrac{1}{y}=-1 \\\dfrac{1}{x+1}= \dfrac{1}{2} \end{cases} \) ⇔ \(\begin{cases} y = - 1\\ x = 1 \end{cases} \)

Vậy hpt có nghiệm duy nhất \(\begin{cases} y = - 1\\ x = 1 \end{cases} \)

Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{1}{y}=\dfrac{-1}{2}\\\dfrac{8}{x+1}+\dfrac{9}{y}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+1}+\dfrac{8}{y}=-4\\\dfrac{8}{x+1}+\dfrac{9}{y}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-1}{y}=1\\\dfrac{1}{x+1}+\dfrac{1}{y}=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(1;-1)

9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)

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

12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)

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

14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)

15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))

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

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

a: =>2/x+2/y=2 và 4/x-2/y=1

=>6/x=3 và 1/x+1/y=1

=>x=2 và 1/y=1-1/2=1/2

=>x=2; y=2

b: Đặt 1/x=a; 1/y=b

=>1/3a+1/3b=1/4 và 5/6a+b=2/3

=>a=1/2; b=1/4

=>x=2; y=4

Ta có: \(\left\{{}\begin{matrix}x-y=10\\\dfrac{300}{y}-\dfrac{300}{x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\dfrac{300}{y}-\dfrac{300}{10+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\dfrac{300\left(y+10\right)}{y\left(y+10\right)}-\dfrac{300y}{y\left(y+10\right)}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\300y+3000-300y=y\left(y+10\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\y^2+10y-3000=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\y^2+10y+25-3025=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\left(y+5\right)^2=3025\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y+10\\x=y+10\end{matrix}\right.\\\left[{}\begin{matrix}y+5=55\\y+5=-55\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=50+10=60\\x=-60+10=-50\end{matrix}\right.\\\left[{}\begin{matrix}y=50\\y=-60\end{matrix}\right.\end{matrix}\right.\)

Vậy: Hệ phương trình có hai cặp nghiệm là (x,y)\(\in\){(-50;-60);(60;50)}

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

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

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

(Các câu khác tương tự nhé.)

a: \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x-35y}{35}=2\\\dfrac{50y-x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-35y=70\\-x+50y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15y=120\\x-35y=70\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=8\\x=70+35y=70+35\cdot8=350\end{matrix}\right.\)

b: ĐKXĐ: x<>0 và y<>0

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{y}=\dfrac{3}{16}-\dfrac{1}{4}=\dfrac{-1}{16}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=48\\\dfrac{1}{x}=\dfrac{1}{16}-\dfrac{1}{48}=\dfrac{2}{48}=\dfrac{1}{24}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=24\\y=48\end{matrix}\right.\left(nhận\right)\)

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

\(\Rightarrow\)Ta có hệ mới: \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\cdot\left(3a-2b\right)=2\cdot4\\3\left(2a+b\right)=3\cdot5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6a-4b=8\left(1\right)\\6a+3b=15 \left(2\right)\end{matrix}\right.\)

Lấy (2)-(1) ta đc:

\(\Rightarrow7b=7\Rightarrow b=1\Rightarrow2a+1=5\Rightarrow a=2\)

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

Với \(x\ne1;y\ne-2\)

hpt <=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\)

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

<=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\2.2+\dfrac{1}{y+2}=5\end{matrix}\right.\)

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

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

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