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a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)(Đk: \(x\ne-1;y\ne-1\))
Đặt \(\dfrac{x}{x+1}\) là A
\(\dfrac{y}{y+1}\) là B
Ta có HPT mới : \(\left\{{}\begin{matrix}2A+B=2\\A+3B=-1\end{matrix}\right.\)(1)
Giải HPT (1) ta được A= \(\dfrac{7}{5}\) ; B=\(-\dfrac{4}{5}\)
+Với A=\(\dfrac{7}{5}\) ta có:
\(\dfrac{x}{x+1}=\dfrac{7}{5}\)
<=>\(5x=7x+7\)
<=>-2x=7
<=> x=\(-\dfrac{7}{2}\)
+Với B = \(-\dfrac{4}{5}\) ta có:
\(\dfrac{y}{y+1}=-\dfrac{4}{5}\)
<=>5y=-4y-4
<=>9y=-4
<=>y=\(-\dfrac{4}{9}\)
Vậy HPT có nghiệm (x;y) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)
Đặt x+y=a; x-2y=b
=>6/a-3/b=3 và 1/a+7/b=2
=>a=5/3 và b=5
=>x+y=5/3 và x-2y=5
=>x=25/9; y=-10/9
Đặt \(\dfrac{x}{\sqrt{4x-1}}=a\)
Theo đề, ta có phương trình:
a+1/a=2
\(\Leftrightarrow a+\dfrac{1}{a}=2\)
\(\Leftrightarrow\dfrac{a^2+1-2a}{a}=0\)
=>a=1
=>\(x=\sqrt{4x-1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4x-1\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=3\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow x\in\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)
a) ĐK : x,y \(\ne0\)
Đặt \(u=\dfrac{1}{x};v=\dfrac{1}{y}\)
Hệ pt đã cho trở thành :
\(\left\{{}\begin{matrix}u-v=1\\3u+4v=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=1+v\\3\left(1+v\right)+4v=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u=1+\dfrac{2}{7}\\v=\dfrac{2}{7}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{9}{7}\\v=\dfrac{2}{7}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y}=\dfrac{2}{7}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}\end{matrix}\right.\)(TM)
Vậy x=7/9 và y=7/2
\(a,ĐK:x,y\ne2\)
Đặt \(\left\{{}\begin{matrix}x-2=a\\y-2=b\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{3}{a}+\dfrac{2}{b}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{a}+\dfrac{9}{b}=15\\\dfrac{6}{a}+\dfrac{4}{b}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{5}{b}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+3=5\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow x=y=3\left(tm\right)\)
\(b,ĐK:x\ge3;y\ge1\)
Sửa: \(\sqrt{x-3}-\sqrt{y-1}=4\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\ge0\\b=\sqrt{y-1}\ge0\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}a-2b=2\\a-b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\-b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3=36\\y-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=39\\y=5\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
Đặt \(\dfrac{1}{y-1}=a\), hpt tở thành
\(\left\{{}\begin{matrix}\dfrac{5}{x+1}+a=10\\\dfrac{1}{x-2}+3a=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{x+1}+3a=30\left(1\right)\\\dfrac{1}{x-1}+3a=18\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\), ta được:
\(\dfrac{15}{x+1}-\dfrac{1}{x-1}=12\\ \Leftrightarrow\dfrac{15x-15-x-1}{\left(x-1\right)\left(x+1\right)}=12\\ \Leftrightarrow12x^2-12=14x-16\\ \Leftrightarrow12x^2-14x+4=0\\ \Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Với \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{10}{3}+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{10y-7}{3\left(y-1\right)}=10\)
\(\Leftrightarrow30y-30=10y-7\Leftrightarrow y=\dfrac{23}{20}\)
Với \(x=\dfrac{2}{3}\Leftrightarrow3+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{1}{y-1}=7\Leftrightarrow7y-7=1\Leftrightarrow y=\dfrac{8}{7}\)
Vậy \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)