a. \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{11\sqrt{x}-5}{x-25}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-\left(11\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-4\sqrt{x}-5+x+5\sqrt{x}-11\sqrt{x}+5}{x-25}=\dfrac{-10\sqrt{x}}{x-25}\)
b. Thay x=4 vào P, ta được
\(P=\dfrac{-10\sqrt{4}}{4-25}=\dfrac{-20}{-21}=\dfrac{20}{21}\)
c. Ta có \(x=51-10\sqrt{26}=26-10\sqrt{26}+25=\left(\sqrt{26}-5\right)^2\)
Thay \(x=\left(\sqrt{26}-5\right)^2\) vào P, ta được
\(P=\dfrac{-10\sqrt{\left(\sqrt{26}-5\right)^2}}{\left(\sqrt{26}-5\right)^2-25}=\dfrac{-10\sqrt{26}+50}{-10\sqrt{26}+26}\)
d. Để \(P=\dfrac{3}{4}\) thì \(\dfrac{-10\sqrt{x}}{x-25}=\dfrac{3}{4}\)
\(\Rightarrow-40\sqrt{x}=3x-75\)
\(\Rightarrow3x+40\sqrt{x}-75=0\)
\(\Rightarrow3x+45\sqrt{x}-5\sqrt{x}-75=0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}+15\right)-5\left(\sqrt{x}+15\right)=0\)
\(\Rightarrow\left(\sqrt{x}+15\right)\left(3\sqrt{x}-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+15=0\\3\sqrt{x}-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-15\\\sqrt{x}=\dfrac{5}{3}\end{matrix}\right.\) \(\Rightarrow x=\dfrac{25}{9}\)
Vậy x thỏa mãn ycbt là \(x=\dfrac{25}{9}\)
\(a,P=\dfrac{\sqrt{x}+1}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{11\sqrt{x}-5}{x-25}\left(x\ge0,x\ne25\right)\\ \Leftrightarrow P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-11\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ \Leftrightarrow P=\dfrac{x-4\sqrt{x}-5+x+5\sqrt{x}-11\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ \Leftrightarrow P=\dfrac{2x-10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ \Leftrightarrow P=\dfrac{2\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+5}\)
\(b,\) Tại \(x=4\)
\(P=\dfrac{2\cdot2}{2+5}=\dfrac{4}{7}\)
\(c,x=51-10\sqrt{26}=\left(\sqrt{26}-5\right)^2\)
Thay vào P, ta được:
\(P=\dfrac{2\left(\sqrt{26}-5\right)}{\sqrt{26}-5+5}=\dfrac{2\sqrt{26}-10}{\sqrt{26}}=\dfrac{52-10\sqrt{26}}{26}=\dfrac{26-5\sqrt{26}}{13}\)
a: Ta có: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{11\sqrt{x}-5}{x-25}\)
\(=\dfrac{x-4\sqrt{x}-5+x+5\sqrt{x}-11\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{2x-10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+5}\)
b: Thay x=4 vào P, ta được:
\(P=\dfrac{2\cdot2}{2+5}=\dfrac{4}{7}\)
