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a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)
a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3=3\)
hay x=0
a: ĐKXĐ: x>0; x<>3
b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c: Để A=-2 thì \(-3\sqrt{x}=4\sqrt{x}+8\)
=>-7 căn x=8(loại)
a: ĐKXĐ: x>0; x<>9
b: \(A=\dfrac{x-3\sqrt{x}-x-9}{x-9}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=-\dfrac{3\sqrt{x}}{2\sqrt{x}+4}\)
c: Để A=-2 thì \(-3\sqrt{x}=-4\sqrt{x}-8\)
=>căn x=-8(loại)
a/ đkxđ: x > 0; x ≠9
b/ \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}:\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c/ \(A=-2\Leftrightarrow\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}=-2\)
\(\Leftrightarrow-3\sqrt{x}=-4\sqrt{x}-8\)
\(\Leftrightarrow\sqrt{x}=-8\) (vô lí)
Vậy không có x nào t/m
Bài 1:
a: \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(x=2+2\sqrt{5}+2-2\sqrt{5}=4\)
Khi x=4 thì \(P=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)
a) điều kiện : \(x>0;x\ne9\)
ta có : \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\) \(\Leftrightarrow A=\left(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\right)\) \(\Leftrightarrow A=\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\) b) \(A< -1\) \(\Leftrightarrow A+1< 0\Leftrightarrow\dfrac{-3\sqrt{x}}{2\sqrt{x}+3}+1< 0\Leftrightarrow\dfrac{-\sqrt{x}+3}{2\sqrt{x}+3}< 0\)\(\Leftrightarrow-\sqrt{x}+3< 0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
vậy ...
----------------------------(-_-)---------------------------------------------------
A=\(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
=\(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x+3}\right)\left(\sqrt{x}-3\right)}\times\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
=\(\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
a, ĐKXĐ: \(x\ge0;x\ne9\)
b, rút gọn
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x}{x-9}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{x-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{x+1}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
c,Cho \(A\le-\dfrac{1}{3}\)
\(< =>\dfrac{3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ < =>\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}\le0\\ < =>\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\le0\\ < =>\dfrac{\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\le0\\ < =>\sqrt{x}-6\le0\\ < =>\sqrt{x}\le36\\ < =>0\le x\le36\)
Vậy để \(A\le-\dfrac{1}{3}\) thì \(0\le x\le36\)và\(x\ne9\)
d, \(A=\dfrac{-3}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}+3\ge3\\ =>\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\\ =>\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{3}\\ =-1\)
Vậy GTNN của A=-1
Xấu ''='' xảy ra khi \(\sqrt{x}=0\\ \Leftrightarrow x=0\)
lần đầu biết San giỏi Toán