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13 tháng 11 2021

\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)

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

5 tháng 7 2021

a) ĐK:\(x\ge0;x\ne9\)

\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b)\(P=-\dfrac{3}{\sqrt{x}+3}\) 

Có \(\sqrt{x}+3\ge3;\forall x\ge0\)

\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)

\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)

a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

b) Để \(A< -\dfrac{1}{3}\) thì \(A+\dfrac{1}{3}< 0\)

\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}< 0\)

\(\Leftrightarrow\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-6< 0\)

\(\Leftrightarrow x< 36\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 36\\x\ne9\end{matrix}\right.\)

3 tháng 6 2023

a) \(P=\left(3-\dfrac{3}{\sqrt{x}-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

\(=\left(\dfrac{3\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\right):\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+2}\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\)

\(=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)

\(=3\sqrt{x}-6\)

b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)   (1)

ĐKXĐ: \(x>0\)

\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)

\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)

\(\Leftrightarrow3x-10\sqrt{x}+1=0\)   (2)

Đặt \(t=\sqrt{x}\ge0\)

\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)

\(\Delta'=25-4=22\)

Phương trình có hai nghiệm phân biệt:

\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)

\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)

Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)

Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)

Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

a: \(P=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=3\sqrt{x}-6\)

b: P=(4căn x-1)/căn x

=>3x-6căn x-4căn x+1=0

=>3x-10căn x+1=0

=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9

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: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)

b: P>=-1/2

=>P+1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)

=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)

=>căn x-9>=0

=>x>=81

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>-6/căn x+3>=-2

Dấu = xảy ra khi x=0