1) \(B=\dfrac{\sqrt{x}-5}{\sqrt{x}}\)
Thay \(x=\dfrac{4}{25}\) vào B, ta được:
\(B=\dfrac{\sqrt{\dfrac{4}{25}}-5}{\sqrt{\dfrac{4}{25}}}\)
\(=\dfrac{\dfrac{2}{5}-5}{\dfrac{2}{5}}\)
\(=\dfrac{-\dfrac{23}{5}}{\dfrac{2}{5}}\)
\(=-\dfrac{23}{2}\)
2) ĐKXĐ: \(x\ne9;x\ge0\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x+9\sqrt{x}}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
3) \(P=A.B\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}.\dfrac{\sqrt{x}-5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}\)
\(=1-\dfrac{2}{\sqrt{x}+3}\)
Để P nhỏ nhất thì \(\dfrac{8}{\sqrt{x}+3}\) lớn nhất
Ta có:
\(\dfrac{8}{\sqrt{x}+3}\ge\dfrac{8}{3}\)
\(\Rightarrow P\) nhỏ nhất là \(1-\dfrac{8}{3}=-\dfrac{5}{3}\) khi \(x=0\)
giup minh cau nay voi
giup minh cau nay voi