Câu 1:
1: Thay x=2,25 vào A, ta được:
\(A=\dfrac{1,5+3}{1,5-5}=\dfrac{4.5}{-3,5}=-\dfrac{7}{9}\)
2: \(B=\dfrac{x+2}{\sqrt{x}}-\dfrac{1}{5-\sqrt{x}}+\dfrac{5x-8\sqrt{x}+10}{x-5\sqrt{x}}\)
\(=\dfrac{x+2}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-5}+\dfrac{5x-8\sqrt{x}+10}{\sqrt{x}\cdot\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\left(x+2\right)\left(\sqrt{x}-5\right)+\sqrt{x}+5x-8\sqrt{x}+10}{\sqrt{x}\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x\sqrt{x}-5x+2\sqrt{x}-10+5x-7\sqrt{x}+10}{\left(\sqrt{x}-5\right)\cdot\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-5\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-5\right)}=\dfrac{x-5}{\sqrt{x}-5}\)
Câu III:
1: ĐKXĐ: x>0 và y<>0
\(\left\{{}\begin{matrix}\sqrt{x}-\dfrac{2}{y}=9\\\sqrt{4x}+\dfrac{7}{y}=51\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-\dfrac{2}{y}=9\\2\sqrt{x}+\dfrac{7}{y}=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}-\dfrac{4}{y}=18\\2\sqrt{x}+\dfrac{7}{y}=51\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{11}{y}=-33\\\sqrt{x}-\dfrac{2}{y}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\\sqrt{x}=9+2:\dfrac{1}{3}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=225\end{matrix}\right.\left(nhận\right)\)
2:
a:
ĐKXĐ: m>=0
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x^2=x\sqrt{m}+m^2+1\)
=>\(\dfrac{1}{2}x^2-x\sqrt{m}-m^2-1=0\)
\(a\cdot c=\dfrac{1}{2}\left(-m^2-1\right)=-\dfrac{1}{2}\left(m^2+1\right)< =-\dfrac{1}{2}< 0\forall m>=0\)
=>(P) luôn cắt (d) tại hai điểm phân biệt
