Câu 1:
ĐKXĐ: x>=0; x<>1
\(P=\dfrac{2}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+1}-\dfrac{5-\sqrt{x}}{x-1}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}-1\right)-5+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{5}{\sqrt{x}+1}\)
Để P=0,25 thì \(\sqrt{x}+1=\dfrac{5}{0,25}=5:\dfrac{1}{4}=20\)
=>\(\sqrt{x}=20-1=19\)
=>\(x=19^2=361\left(nhận\right)\)
Câu 2:
\(\left\{{}\begin{matrix}5-2\left(x+y\right)=-3y\\x-1=2y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5-2x-2y+3y=0\\x-2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x+y=-5\\x-2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-5\\x-2\left(2x-5\right)=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-5\\x-4x+10=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-6\\y=2x-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\cdot2-5=4-5=-1\end{matrix}\right.\)
\(T=2025\cdot x_0-2026\cdot y_0\)
\(=2025\cdot2-2026\cdot\left(-1\right)=4050+2026=6076\)
