Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: x # -1/2; y # -2
\(Đặt\ \dfrac{x-1}{2x+1}=a; \dfrac{y-2}{y+2}=b \\Hệ\ tương\ đương: \\\begin{cases} a-b=1\\3a+2b=3 \end{cases} <=> \begin{cases} 3a-3b=3\\3a+2b=3 \end{cases} \\<=>\begin{cases} -5b=0\\a-b=1 \end{cases} <=>\begin{cases} b=0\\a=1 \end{cases} \\->\begin{cases} x-1=2x+1\\y-2=0 \end{cases} <=>\begin{cases} x=-2(thoả\ ĐKXĐ)\\y=2(thoả\ ĐKXĐ) \end{cases}\)
b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy-x-y+1=6\\\dfrac{1}{\left(x-1\right)^2-1}+\dfrac{1}{\left(y-1\right)^2-1}=\dfrac{2}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-1\right)-\left(y-1\right)=6\\\dfrac{1}{\left(x-1\right)^2-1}+\dfrac{1}{\left(y-1\right)^2-1}=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=6\\\dfrac{1}{\left(x-1\right)^2-1}+\dfrac{1}{\left(y-1\right)^2-1}=\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\) đặt \(\left\{{}\begin{matrix}x-1=a\\y-1=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a.b=6\Rightarrow b=\dfrac{6}{a}\\\dfrac{1}{a^2-1}+\dfrac{1}{b^2-1}=\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{a^2-1}+\dfrac{1}{\dfrac{36}{a^2}-1}=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{a^2-1}+\dfrac{a^2}{36-a^2}=\dfrac{2}{3}\Rightarrow a^4-16a^2+36=0\)
\(\Rightarrow\left[{}\begin{matrix}a^2=8+2\sqrt{7}\\a^2=8-2\sqrt{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\pm\sqrt{8+2\sqrt{7}}=\pm\left(\sqrt{7}+1\right)\\a=\pm\sqrt{8-2\sqrt{7}}=\pm\left(\sqrt{7}-1\right)\end{matrix}\right.\)
\(x=a+1\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{7}\\x=-\sqrt{7}\\x=\sqrt{7}\\x=2-\sqrt{7}\end{matrix}\right.\) \(\Rightarrow y=\dfrac{6}{a}+1=\left[{}\begin{matrix}\sqrt{7}\\2-\sqrt{7}\\2+\sqrt{7}\\-\sqrt{7}\end{matrix}\right.\)
Vậy hệ đã cho có 4 cặp nghiệm thỏa mãn:
\(\left(x;y\right)=\left(2+\sqrt{7};\sqrt{7}\right)\),\(\left(-\sqrt{7};2-\sqrt{7}\right)\),\(\left(\sqrt{7};2+\sqrt{7}\right)\) ,\(\left(2-\sqrt{7};-\sqrt{7}\right)\)
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) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)
phương trình 2 ⇔\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}=7-3xy\)⇔\(\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2=7-3xy\)
đoạn sau bạn tự giải nha