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\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (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={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
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..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
a: Đặt 1/x=a; 1/y=b
Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}3a+5b=-\dfrac{3}{2}\\5a-2b=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
b: Đặt \(\dfrac{1}{x+y-1}=a;\dfrac{1}{x-y+1}=b\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a-4b=\dfrac{-14}{5}\\3a+2b=-\dfrac{13}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-1=-1\\x-y+1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5
Tất cả các hpt này đều giải bằng PP đặt ẩn phụ
a) \(\begin{cases}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{cases}\)
Đặt \(x+y=a\) ; \(x-y=b\) ta được:
\(\begin{cases}2a+3b=4\\a+2b=5\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}2a+3b=4\\2a+4b=10\end{cases}\)\(\Leftrightarrow\) \(\begin{cases}-b=-6\\2a+4b=10\end{cases}\)
\(\Leftrightarrow\) \(\begin{cases}b=6\\2a+4.6=10\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}a=-7\\b=6\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}x+y=6-7\\x-y=6-7\end{cases}\)
\(\Leftrightarrow\) \(\begin{cases}x-7=-1\\6-y=-1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}x=6\\y=-7\end{cases}\)
Lúc khác mình làm tiếp mấy câu kia
Tiếp nào!
b) \(\begin{cases}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{cases}\) Đặt \(\dfrac{1}{x}=a\) ; \(\dfrac{1}{y}=b\) ta được:
\(\begin{cases}3a-4b=2\\4a-5b=3\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}12a-16b=8\\12a-15b=9\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}-1b=-1\\12a-15b=9\end{cases}\)
\(\Leftrightarrow\) \(\begin{cases}b=1\\a=2\end{cases}\)\(\Leftrightarrow\) \(\begin{cases}a=2\\b=1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}\dfrac{1}{a}=2\\\dfrac{1}{b}=1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}a=\dfrac{1}{2}\\b=1\end{cases}\)
c) Làm tương tự thay \(\dfrac{1}{2x-y}=a\) ; \(\dfrac{1}{x+y}=b\)
a, Let's \(\dfrac{1}{x+1}=a;\dfrac{1}{y-1}=b\), we have:
\(\left\{{}\begin{matrix}3a+b=2\\2a-3b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3a\\2a-3\left(2-3a\right)=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3a\\2a-6+9a=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3a\\11a=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3\cdot1\\a=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=-1\\\dfrac{1}{y-1}=1\end{matrix}\right.\)(remember \(\left(x;y\right)\ne-1;1\) :>)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=-1\\y-1=1\end{matrix}\right.\\ \left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\) (satisfied)
So equations (i don't know word "hệ phương trình" in English :>) have 1 root \(\left(x;y\right)=\left(-2;2\right)\).
Enjoy
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a: \(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{1}{y-1}=2\\\dfrac{2}{x+1}-\dfrac{3}{y-1}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+1}+\dfrac{3}{y-1}=6\\\dfrac{2}{x+1}-\dfrac{3}{y-1}=5\end{matrix}\right.\)
=>11/x+1=11 và 1/y-1=2-3/x+1
=>x+1=1 và 1/y-1=2-3=-1
=>x=0; y-1=-1
=>x=0; y=0
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{6}{y-1}=2\\\dfrac{2}{x+1}+\dfrac{4}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{10}{y-1}=-1\\\dfrac{1}{x+1}=1+\dfrac{3}{y-1}\end{matrix}\right.\)
=>y-1=10; 1/x+1=1+3/10=13/10
=>y=11; x=10/13-1=-3/13
Bài 1:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}=1\\\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)
Bài 2:
Theo đề, ta có:
\(\left\{{}\begin{matrix}2a-3b=4\\-a-2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-3b=4\\-2a-4b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{12}{7}\\a=-\dfrac{4}{7}\end{matrix}\right.\)