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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={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
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:
\(\left\{{}\begin{matrix}2a-4b=\dfrac{-14}{5}\\3a+2b=-\dfrac{13}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-1=-1\\x-y+1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
c: =>3x^2+3y^2=39 và 3x^2-2y^2=-6
=>5y^2=45 và x^2=13-y^2
=>y^2=9 và x^2=4
=>\(\left\{{}\begin{matrix}x\in\left\{2;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x}=5\\\sqrt{x}-\sqrt{y}=-\dfrac{11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y}=1+\dfrac{11}{2}=\dfrac{13}{2}\end{matrix}\right.\)
=>x=1 và y=169/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4-3=1\\-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9-2=7\end{matrix}\right.\)
=>x+1=11/9 và y+4=-11/19
=>x=2/9 và y=-87/19
a) \(x^4-x^2-2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-2\right)=0\)
\(\Leftrightarrow...\)
a)\(x^4-x^2-2=0\)
\(\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+1\right)=0\)
Dễ thấy: \(x^2+1\ge1>0 \forall x\) (loại)
\(\Rightarrow x^2-2=0\Rightarrow x^2=2\Rightarrow x=\pm\sqrt{2}\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{3}{4}\\\dfrac{1}{x-y}-\dfrac{1}{x+y}=\dfrac{1}{4}\end{matrix}\right.\)
Cộng theo vế 2 pt ta có:
\(\dfrac{1}{x+y}+\dfrac{1}{x-y}+\dfrac{1}{x-y}-\dfrac{1}{x+y}=1\)
\(\Leftrightarrow\dfrac{1}{x-y}+\dfrac{1}{x-y}=1\Leftrightarrow\dfrac{2}{x-y}=1\Leftrightarrow x-y=2\)
Trừ theo vế 2 pt ta có:
\(\dfrac{1}{x+y}+\dfrac{1}{x-y}-\dfrac{1}{x-y}+\dfrac{1}{x+y}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{x+y}+\dfrac{1}{x+y}=\dfrac{1}{2}\)\(\Leftrightarrow\dfrac{2}{x+y}=\dfrac{1}{2}\Leftrightarrow x+y=4\)
Ta có hpt \(\left\{{}\begin{matrix}x-y=2\\x+y=4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
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
a) ĐKXĐ: \(x\ne0,\text{ }y\ne0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\), hệ phương trình đã cho trở thành:
\(\left\{{}\begin{matrix}3a+5b=\dfrac{-3}{2}\\5a-2b=\dfrac{8}{3}\end{matrix}\right.\)
Giải hệ này bằng phương pháp cộng đại số hoặc thế tìm được 1 nghiệm duy nhất: \(\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{-1}{2}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=\dfrac{-1}{2}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)(thỏa mãn ĐKXĐ)
Vậy hệ đã cho có 1 nghiệm duy nhất \(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\).