\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{2}{3}\\\dfrac{1}{4x}+\dfrac{1}{3y}=\dfrac{1}{5}\end{matrix}\right.\) (II)
Đặt: \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\left(x\ne0\right)\\b=\dfrac{1}{y}\left(y\ne0\right)\end{matrix}\right.\)
(II) <=> \(\left\{{}\begin{matrix}a+b=\dfrac{2}{3}\\\dfrac{1}{4}a+\dfrac{1}{3}b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{4}{15}\\b=\dfrac{2}{5}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=\dfrac{4}{15}=\dfrac{1}{x}\\b=\dfrac{2}{5}=\dfrac{1}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{15}{4}\left(TM\right)\\y=\dfrac{5}{2}\left(TM\right)\end{matrix}\right.\)
Vậy: Hpt (II) có nghiệm (x;y)= (15/4;5/2)
Câu 1:
a) ĐKXĐ: \(x\ge-1\)
Ta có: \(\sqrt{2x+5}-1=x\)
\(\Leftrightarrow\sqrt{2x+5}=x+1\)
\(\Leftrightarrow2x+5=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-2x-5=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Vậy: S={2}
