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Nếu đề là y+1 thì
\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\2+\dfrac{2}{x-2}-1-\dfrac{1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+2b=\dfrac{17}{5}\\2a-b=\dfrac{21}{5}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\2+\dfrac{2}{x-1}-1-\dfrac{3}{y-1}=\dfrac{26}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2}+\dfrac{4}{y+1}=\dfrac{34}{5}\\\dfrac{6}{x-1}-\dfrac{9}{y-1}=\dfrac{63}{5}\end{matrix}\right.\)
\(\dfrac{4}{y+1}+\dfrac{9}{y-1}=-\dfrac{29}{5}=>y=....\)
Sửa đề: \(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-2}{x-2}+\dfrac{y+2}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-4+2}{x-2}+\dfrac{y+1+1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=\dfrac{11}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=1\\\dfrac{1}{y+1}=\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y+1=5\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3;4\right)\)
\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-2}{x-2}+\dfrac{y+1}{y-1}=\dfrac{26}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{y+1}=\dfrac{17}{5}-\dfrac{3}{x-2}\\\dfrac{2x-2}{x-2}+\dfrac{y-1+2}{y-1}=\dfrac{26}{5}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{y+1}=\dfrac{17}{5}-\dfrac{3}{x-2}\\\dfrac{2x-2}{x-2}+1+\dfrac{2}{y-1}=\dfrac{26}{5}\end{matrix}\right.\left\{{}\begin{matrix}\dfrac{2}{y+1}=.......\\\dfrac{2}{y-1}=\dfrac{21}{5}-\dfrac{2x-2}{x-2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{17}{5}-\dfrac{3}{x-2}=\dfrac{21}{5}-\dfrac{2x-2}{x-2}\)\(\Rightarrow\dfrac{4}{5}=\dfrac{2x-5}{x-2}\Rightarrow10x-25=4x-8\Rightarrow x=\dfrac{17}{6}\Rightarrow y=-11\)
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
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: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x+4-5}{x+2}-\dfrac{5}{y-1}=-\dfrac{14}{3}\\\dfrac{3}{x+2}+\dfrac{2y-2+5}{y-1}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{x+2}-\dfrac{5}{y-1}=-\dfrac{14}{3}-2=-\dfrac{20}{3}\\\dfrac{3}{x+2}+\dfrac{5}{y-1}=6\end{matrix}\right.\)
=>x+2=3 và y-1=1
=>x=1 và y=2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2x}{x-1}+\dfrac{3}{y+2}=\dfrac{-2}{5}\\\dfrac{-5}{x-1}-\dfrac{4y}{y+2}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2x+2-2}{x-1}+\dfrac{3}{y+2}=\dfrac{-2}{5}\\\dfrac{-5}{x-1}-\dfrac{4y+8-8}{y+2}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{x-1}+\dfrac{3}{y+2}=-\dfrac{2}{5}+2=\dfrac{8}{5}\\\dfrac{-5}{x-1}+\dfrac{8}{y+2}=\dfrac{1}{10}-4=-\dfrac{39}{10}\end{matrix}\right.\)
=>x-1=-2/49 và y+2=-5/79
=>x=47/49 và y=-5/79-2=-163/79
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
ĐK: \(x>\dfrac{1}{4};y>-\dfrac{1}{4}\), đặt \(a=\dfrac{1}{\sqrt{12x-3}};b=\dfrac{1}{\sqrt{4y+1}}\)với a,b>0
khi đó, ta có hệ phương mới \(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}80a+40b=8\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45a=3\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35.\dfrac{1}{15}+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\b=\dfrac{1}{15}\end{matrix}\right.\)
thay \(\dfrac{1}{\sqrt{12x-3}}=a\) hay \(\dfrac{1}{\sqrt{12x-3}}=\dfrac{1}{15}\Rightarrow\sqrt{12x-3}=15\Leftrightarrow12x-3=225\Leftrightarrow12x=228\Leftrightarrow x=19\left(TMĐK\right)\) thay \(\dfrac{1}{\sqrt{4y+1}}=b\) hay
\(\dfrac{1}{\sqrt{4y+1}}=\dfrac{1}{15}\Rightarrow\sqrt{4y+1}=15\Leftrightarrow4y+1=225\Leftrightarrow4y=224\Leftrightarrow y=56\left(TMĐK\right)\)
Vậy (x;y)=(9;56) là nghiệm duy nhất của hệ phương trình đã cho.
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
ĐK: x,y#0, khi đó \(\dfrac{1}{x}+\dfrac{1}{y}=4\Rightarrow x+y=4xy\)
Do đó \(x\left(1+4y\right)+y=2\Leftrightarrow x+4xy+y=2\Leftrightarrow x+x+y+y=2\Leftrightarrow2\left(x+y\right)=2\Leftrightarrow x+y=1\)
Mà \(4xy=x+y\Leftrightarrow4xy=1\Leftrightarrow xy=\dfrac{1}{4}\)
Vậy \(x+y=1;xy=\dfrac{1}{4}\)
Do đó x,y là nghiệm của phương trình:
\(t^2-t+\dfrac{1}{4}=0\)
\(\Delta=b^2-4ac=1-4.1.\dfrac{1}{4}=0\)
Phương trình có nghiêm kép \(x_1=x_2=-\dfrac{b}{2a}=-\dfrac{-1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=y=\dfrac{1}{2}\left(nhận\right)\)
Vậy (x;y)=\(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) là nghiệm duy nhất của hệ phương trình đã cho.
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2x-4+2}{x-2}+\dfrac{y+1+1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=\dfrac{26}{5}-1-2=\dfrac{11}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y+1=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-\dfrac{4}{5}\end{matrix}\right.\)