\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0\)

\(\Leftrightarrow\dfrac{5}{5\sqrt{x}+10}-\dfrac{\sqrt{x}+2}{5\sqrt{x}+10}>0\)

\(\Leftrightarrow\dfrac{5-\sqrt{x}-2}{5\sqrt{x}+10}>0\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-3\right)}{5\sqrt{x}+10}>0\)

Mà: \(5\sqrt{x}+10\ge10>0\forall x\)

\(\Leftrightarrow\sqrt{x}>3\)

\(\Leftrightarrow x>9\)

_________

\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{4}{2\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+6}< 0\)

\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\sqrt{x}+6}< 0\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)}{2\sqrt{x}+6}< 0\)

Mà: \(2\sqrt{x}+6\ge6>0\forall x\)

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

\(\Leftrightarrow\sqrt{x}< 1\)

\(\Leftrightarrow x< 1\)

\(\Leftrightarrow0\le x\le1\)

1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)

=>4x+8=3x-1

=>x=-9

2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)

=>8x-4=5x-7

=>3x=-3

=>x=-1

3: ĐKXD: x>=0

\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

=>\(x+\sqrt{x}-6=x-1\)

=>căn x=-1+6=5

=>x=25

4: ĐKXĐ: x>=0

PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

=>x-2*căn x-3=x-4

=>-2căn x-3=-4

=>2căn x+3=4

=>2căn x=1

=>căn x=1/2

=>x=1/4

1: ĐKXĐ: \(-1< x< 1\)

2: ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)

3: ĐKXĐ: \(\left[{}\begin{matrix}x< -3\\x\ge2\end{matrix}\right.\)

4: ĐKXĐ: \(2< a\le3\)

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

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

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

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

\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)

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

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

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

a: Ta có: \(2\sqrt{2}-\dfrac{1}{2}\cdot\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\cdot\dfrac{1}{2}=2\sqrt{2}\)

\(\Leftrightarrow\sqrt{x}=4\sqrt{2}\)

hay x=32

b: Ta có: \(2\sqrt{x}-\sqrt{\dfrac{x}{3}}=1\)

\(\Leftrightarrow2\sqrt{x}-\dfrac{\sqrt{3}}{3}\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\dfrac{6+\sqrt{3}}{11}\)

hay \(x=\dfrac{39+12\sqrt{3}}{121}\)

c: Ta có: \(4\sqrt{x}+\sqrt{\dfrac{x}{2}}=\dfrac{1}{3}\)

\(\Leftrightarrow4\sqrt{x}+\dfrac{\sqrt{2}}{2}\sqrt{x}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{x}=\dfrac{8-\sqrt{2}}{93}\)

hay \(x=\dfrac{66-16\sqrt{2}}{8649}\)

Ta có: \(P=A\cdot B\) (ĐK: \(x>0;x\ne4\))

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

\(=\left[\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)

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

\(=\left(1+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+9}\)

Với x > 0; x ≠ 4 thì \(\sqrt{P}< \dfrac{1}{3}\Leftrightarrow P< \dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}< \dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}-\dfrac{1}{9}< 0\)

\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-1\right)}{9\left(\sqrt{x}+9\right)}-\dfrac{\sqrt{x}+9}{9\left(\sqrt{x}+9\right)}< 0\)

\(\Leftrightarrow\dfrac{9\sqrt{x}-9-\sqrt{x}-9}{9\sqrt{x}+81}< 0\)

\(\Leftrightarrow\dfrac{8\sqrt{x}-18}{9\sqrt{x}+18}< 0\)

Ta thấy: \(9\sqrt{x}+18>0\forall x\)

\(\Rightarrow8\sqrt{x}-18< 0\)

\(\Rightarrow\sqrt{x}< \dfrac{18}{8}\)

\(\Rightarrow\sqrt{x}< \dfrac{9}{4}\Leftrightarrow x< \dfrac{81}{16}\)

Kết hợp với điều kiện, ta được: \(0< x\le5\)\(;x\ne4\)

\(\Rightarrow x\in\left\{1;2;3;5\right\};x\in Z\) thì \(\sqrt{P}< \dfrac{1}{3}\)

#Urushi

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

\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)

=2

Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)

a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)

b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)

=>x<1

=>x<1

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

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

\(=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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