\(\dfrac{\text{1}}{\text{x}-5}+\dfrac{6}{\sqrt{y}+2}=2\)
⇔ \(\dfrac{\text{2}}{\text{x}\text{ }-5}+\dfrac{12}{\sqrt{y}+2}=4\)
⇒ \(\dfrac{\text{13}}{\sqrt{\text{y}\text{ }}+2}=\text{13}\)
⇔ \(\sqrt{\text{y}}+2=1\)
⇒ \(\sqrt{\text{y}}=-\text{1}\)
Vì \(\sqrt{\text{y}}\) ≥ 0 ∀ y
⇒ phương trình vô nghiệm
giải các hpt sau: a)\(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{3x}{4}+\dfrac{2y}{5}=2,3\\x-\dfrac{3y}{5}=0,8\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)cíu zới
GIẢI HPT
\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{6}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{\left|x-1\right|}+\dfrac{3}{\sqrt{y-2}}=4\\\dfrac{5}{\left|x-1\right|}-\dfrac{2}{\sqrt{y-2}}=3\end{matrix}\right.\)
Bài 1: Giải các hệ PT
a) \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+1\right)=\dfrac{1}{2}xy+5\\\dfrac{1}{3}\left(x-3\right)\left(y-5\right)=\dfrac{1}{3}xy-\dfrac{4}{3}\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{3}{\sqrt[]{x+y}}-\dfrac{2}{\sqrt[]{x-y}}=4\\\dfrac{2}{\sqrt[]{x+y}}+\dfrac{1}{\sqrt[]{x-y}}=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x+2}+2\sqrt{y-1}=5\\\dfrac{2}{x+2}+3\sqrt{y-1}=8\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=\dfrac{5}{2}\\x+y=5\end{matrix}\right.\)
giải hệ: \(\left\{{}\begin{matrix}\dfrac{1}{x-y}+\dfrac{1}{x+y}=2\\\dfrac{2}{x+y}+\dfrac{3}{x+y}=5\end{matrix}\right.\)
giải pt: \(\sqrt{x^2-4x+7}=\sqrt{x+1}\)
\(\left\{{}\begin{matrix}\dfrac{3}{\left|x+2\right|}+\dfrac{1}{\sqrt{y-2}}=4\\\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\)
