b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)

Thay vào (1) ta được :

\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}

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

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = 1/x ; b = 1/y

Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)

Vậy S = {(\(\dfrac{3}{4};-3\) )}

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

ĐK xác định : x≠0 ; y ≠0

Áp dụng quy tác cộng đại số ta có :

\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}

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

ĐK xác định : x≠0 ; y≠0

áp dụng quy tắc cộng đại số ta có :

\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)

Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}

e) ĐK xác định x≠0 ; y≠0

\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)

Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}

a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)

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

b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)

Phương trình đâu bạn ?

$x=144,$ $y=36$.

a.

1x;b=1y.

⎧⎪ ⎪⎨⎪ ⎪⎩a+b=56
2416.a+15.b=215⇔{a=12b=14

⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1x=121y=13

⇔{x=2y=3
c,

xy−xy+12=1xy−12−xy=2

xy=a;xy+12=b;xy−12=c

1b+1c=2a(1)

1b−1a)+(1c−1a)=0

a−bab+a−cac=0⇔1ab−2ac=0

1a(1b−2c)=0⇒1b=2c⇒c=2b

$\Rightarrow \{\begin{array}{c}a-b=1\\ 2b-a=2\end{array}\iff \{\begin{array}{c}a=4\\ b=3\end{array}$

⇒⎧⎪ ⎪⎨⎪ ⎪⎩xy=4xy+12=3⇔⎧⎪⎨⎪⎩x=4y

Giải hệ sau :

Câu a :

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

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

Vậy ...........................

Câu b :

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) . Ta có :

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{3}{5}\\3a+4b=2\end{matrix}\right.\)

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

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

Vậy..................

\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)

hỏi trước tí, bạn biết giải cái hệ này chứ?

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

ba cái đồ êu!!

câu số 6 (con số của quỷ sa tăng :v)

đặt \(\left\{{}\begin{matrix}a=\left|x\right|\\b=\left|y\right|\end{matrix}\right.\) (a,b >/ 0)

hpt trở thành : \(\left\{{}\begin{matrix}a+4b=18\\3a+b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x\right|=2\\\left|y\right|=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=4\\y=-4\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt có các ng (x;y) là: (có 4 nghiệm tự kết luận)

1, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\) (I) (ĐKXĐ: x, y \(\ne\)0)

Đặt \(\dfrac{1}{x}=a\) ; \(\dfrac{1}{y}=b\)

Hệ pt (I) trở thành :

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