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Trả lời:
a, \(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\left(ĐK:x\ge0;x\ne25\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b, Thay x = 9 vào A, ta được:
\(A=\frac{\sqrt{9}-5}{\sqrt{9}+5}=\frac{3-5}{3+5}=\frac{-2}{8}=-\frac{1}{4}\)
c, \(A< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{3\left(\sqrt{x}-5\right)}{3\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-15-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-20}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Rightarrow2\sqrt{x}-20< 0\) (vì \(3\left(\sqrt{x}+5\right)>0\) )
\(\Leftrightarrow2\sqrt{x}< 20\)
\(\Leftrightarrow\sqrt{x}< 10\)
\(\Leftrightarrow x< 100\)
Vậy \(0\le x< 100\)và \(x\ne25\) là giá trị cần tìm.
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
a: \(P=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b: Khi x=9 thì \(P=\dfrac{3-5}{3+5}=\dfrac{-2}{8}=\dfrac{-1}{4}\)
c: Để P=1/2 thì căn x-5/căn x+5=1/2
=>2 căn x-10=căn x+5
=>căn x=15
=>x=225
Sửa đề chút nhé
Đk: x khác 25, x lớn bằng 0
\(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{x-25}-\frac{10\sqrt{x}}{x-25}-\frac{5\left(\sqrt{x}-5\right)}{x-25}\)
=\(\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b) Em tự làm
c) với đk trên
\(\frac{\sqrt{x}-5}{\sqrt{x}+5}< \frac{1}{3}\Leftrightarrow3\sqrt{x}-15< \sqrt{x}+5\Leftrightarrow2\sqrt{x}< 20\Leftrightarrow x< 100\)
Vậy \(0\le x\le100,x\ne25\)
thank c nha