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

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-\left(a-4\right)}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

b: P>1/6

=>P-1/6>0

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

=>\(\dfrac{6\left(\sqrt{a}-2\right)-3\sqrt{a}}{18\sqrt{a}}>0\)

=>\(6\left(\sqrt{a}-2\right)-3\sqrt{a}>0\)

=>\(3\sqrt{a}-12>0\)

=>\(\sqrt{a}>4\)
=>a>16

16¹⁹ và 8²⁵

16¹⁹ = (2⁴)¹⁹ = 2⁷⁶

8²⁵ = (2³)²⁵ = 2⁷⁵

Vì 2⁷⁵ < 2⁷⁶ nên 

16¹⁹ > 8²⁵

a: \(A=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt[]{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}+3}=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)

\(=\dfrac{4x}{\sqrt{x}-3}\)

b: A=-2

=>\(4x=-2\left(\sqrt{x}-3\right)=-2\sqrt{x}+6\)

=>\(4x+2\sqrt{x}-6=0\)

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

=>\(\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

mà \(2\sqrt{x}+3>=3>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1=0\)

=>x=1(nhận)

a: \(Q=\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)

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

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

\(P=\dfrac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)

\(=\dfrac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-2}\)

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

b: P=Q

=>\(x-1=2\sqrt{x}+1\)

=>\(x-2\sqrt{x}-2=0\)

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

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

mà \(\sqrt{x}-1>=-1\) với mọi x thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1=\sqrt{3}\)

=>\(\sqrt{x}=1+\sqrt{3}\)

=>\(x=\left(1+\sqrt{3}\right)^2=4+2\sqrt{3}\left(nhận\right)\)

a: \(P=\dfrac{3}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-5}{x-1}\)

\(=\dfrac{3}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\left(\sqrt{x}-1\right)-\sqrt{x}-1-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b: \(x=24-16\sqrt{2}=8\left(3-2\sqrt{2}\right)=8\left(\sqrt{2}-1\right)^2\)

Thay \(x=8\left(\sqrt{2}-1\right)^2\) vào P, ta được:

\(P=\dfrac{1}{\sqrt{8\left(\sqrt{2}-1\right)^2}-1}\)

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

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

ĐKXĐ: x>=0; x<>4

a: Thay x=9 vào A, ta được:

\(A=\dfrac{3}{3-2}=\dfrac{3}{1}=3\)

b: T=A-B

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}-\dfrac{4\sqrt{x}}{x-4}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-2\left(\sqrt{x}-2\right)-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

c: Để T nguyên thì \(\sqrt{x}-2⋮\sqrt{x}+2\)

=>\(\sqrt{x}+2-4⋮\sqrt{x}+2\)

=>\(-4⋮\sqrt{x}+2\)

mà \(\sqrt{x}+2>=2\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}+2\in\left\{2;4\right\}\)

=>\(x\in\left\{0;4\right\}\)

Kết hợp ĐKXĐ, ta được: x=0

a: Thay x=9 vào P, ta được:

\(P=\dfrac{9+3}{\sqrt{9}-2}=\dfrac{12}{3-2}=\dfrac{12}{1}=12\)

b: \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

c: Đặt A=P:Q

\(=\dfrac{x+3}{\sqrt{x}-2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{3}{\sqrt{x}}}=2\sqrt{3}\) với mọi x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=3\)

=>x=3(nhận)

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

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

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

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

b: \(2P=2\sqrt{x}+5\)

=>\(2\left(\sqrt{x}+1\right)=\sqrt{x}\left(2\sqrt{x}+5\right)\)

=>\(2x+5\sqrt{x}-2\sqrt{x}-2=0\)

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

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

mà \(\sqrt{x}+2>=2>0\forall x\) thỏa mãn ĐKXĐ

nên \(2\sqrt{x}-1=0\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>\(x=\dfrac{1}{4}\left(nhận\right)\)

a) Do x ⋮ 5

⇒ x ∈ B(5) = {0; 5; 10; ...; 35; 40; 45; ...}

Mà x ≤ 40

⇒ x ∈ {0; 5; 10; ...; 35; 40}

b) Do 16 ⋮ x

⇒ x ∈ Ư(16) = {1; 2; 4; 8; 16}

c) 2x + 3 ∈ Ư(10) = {1; 2; 5; 10}

Mà x là số tự nhiên nên 2x + 3 ≥ 3 và 2x + 3 là số tự nhiên lẻ

⇒ 2x + 3 ∈ {5}

⇒ 2x ∈ {2}

⇒ x ∈ {1}