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17 tháng 10 2021
\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
30 tháng 8 2021
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b: Để \(A>\dfrac{1}{3}\) thì \(A-\dfrac{1}{3}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)
30 tháng 8 2021
a) \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne0\right)\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\)
b) \(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\)
\(\Leftrightarrow6>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)
31 tháng 8 2023
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
18 tháng 9 2021
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
18 tháng 9 2021
a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
29 tháng 10 2021
\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(=\dfrac{-3}{2\sqrt{x}+4}\)
Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)
\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)
hay \(0\le x< \dfrac{25}{4}\)
8 tháng 11 2021
a: \(Q=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
3 tháng 12 2022
a: Khi x=16 thì B=4+1=5
b: \(A=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)
HP
3 tháng 9 2021
a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
3 tháng 9 2021
\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)
Mà \(\sqrt{x}\ge0\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy \(x=0\) thì A nguyên
11 tháng 12 2023
|P|+P=0
=>|P|=-P
=>P<=0
=>\(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}< =0\)
=>\(\sqrt{x}-3< =0\)
=>\(\sqrt{x}< =3\)
=>0<=x<=9
kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 9\\x< >4\end{matrix}\right.\)
mà x là số nguyên tố
nên \(x\in\left\{2;3;5;7\right\}\)
Với x > = 0
a, Ta có : √x+3≥3⇒P=3√x+3≤33=1x+3≥3⇒P=3x+3≤33=1
Dấu ''='' xảy ra khi x = 0
Vậy GTLN của A bằng 1 tại x = 0