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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 ≠0≠0
Đặt u=1x;v=1yu=1x;v=1y
Hệ pt đã cho trở thành :
{u−v=13u+4v=5{u−v=13u+4v=5
⇔{u=1+v3(1+v)+4v=5⇔{u=1+v3(1+v)+4v=5⇔⎧⎪ ⎪⎨⎪ ⎪⎩u=1+27v=27⇔{u=1+27v=27⇔⎧⎪ ⎪⎨⎪ ⎪⎩u=97v=27⇔{u=97v=27 ⇔⎧⎪ ⎪ ⎪
a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
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) \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)(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) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)
\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)
\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)
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 \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)