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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:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
1: \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-3\sqrt{5}+3\cdot3\sqrt{2}+6\sqrt{2}\)
\(=-\sqrt{5}+15\sqrt{2}\)
2:
a: \(A=\left(\dfrac{3\sqrt{x}-7}{x-\sqrt{x}-6}-\dfrac{3\sqrt{x}-6}{x-4}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}+2}{x+4\sqrt{x}+4}\)
\(=\left(\dfrac{3\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{3}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2}\)
\(=\dfrac{3\sqrt{x}-7-3\left(\sqrt{x}-3\right)+\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{4\sqrt{x}-5-3\sqrt{x}+9}{\sqrt{x}-3}=\dfrac{\sqrt{x}+4}{\sqrt{x}-3}\)
b: A<1
=>A-1<0
=>\(\dfrac{\sqrt{x}+4-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>\(\dfrac{7}{\sqrt{x}-3}< 0\)
=>\(\sqrt{x}-3< 0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: 0<=x<9 và x<>4