6: Để P>1 thì P-1>0

\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}>0\)

\(\Leftrightarrow\sqrt{a}-2< 0\)

hay a<4

Kết hợp ĐKXĐ, ta được: \(0\le a< 4\)

5: Để P>0 thì \(x-4\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}-4>0\)

hay x>16

1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

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

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

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

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

\(\Leftrightarrow x=1\left(nhận\right)\)

2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)

\(b,M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ x=3+2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}+1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)=1\\ c,M>0\Leftrightarrow\sqrt{x}-2>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>4\)

Phần a,b,c bạn có thể tham khảo bài bên dưới. 

Phần d.

ĐKXĐ: $x\geq 0; x\neq 4$

$A>5\Leftrightarrow \frac{x+9}{2\sqrt{x}}>5$ ($x> 0$)

$\Leftrightarrow x+9> 10\sqrt{x}$

$\Leftrightarrow x-10\sqrt{x}+9>0$

$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-9)>0$

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} \sqrt{x}-1>0\\ \sqrt{x}-9>0\end{matrix}\right.\\ \left\{\begin{matrix} \sqrt{x}-1<0\\ \sqrt{x}-9<0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>1\\ x>81\end{matrix}\right.\\ \left\{\begin{matrix} 0\leq x< 1\\ 0\leq x< 81\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x>81\\ 0\leq x< 1\end{matrix}\right.\)

Kết hợp với đkxđ suy ra $x>81$ hoặc $0< x< 1$

a

Với: x \(\ge0,x\) \(\ne4\) có:

\(A=\left(\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right):\left(\dfrac{\left(\sqrt{x}+2\right)^2}{x-4}-\dfrac{\left(\sqrt{x}-2\right)^2}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)

\(=\left(\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4}{x-4}-\dfrac{x-4\sqrt{x}+4}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)

\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-6\sqrt{x}}{x-4}\right)\)

\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{2\sqrt{x}}{x-4}\right)\)

\(=\dfrac{\left(x+9\right)\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}=\dfrac{x+9}{2\sqrt{x}}\)

b

Giải \(x^2-5x+4=0\)

Nhẩm nghiệm: a + b + c = 0 (1 - 5 + 4 = 0)

\(\Rightarrow x_1=1;x_2=\dfrac{c}{a}=\dfrac{4}{1}=4\)

Thay x = 1 vào A:

\(A=\dfrac{1+9}{2\sqrt{1}}=\dfrac{10}{2}=5\)

Thay x = 4 vào A:

\(A=\dfrac{4+9}{2.\sqrt{4}}=\dfrac{13}{2.2}=\dfrac{13}{4}\)

c

ĐK: x > 0

\(A=0\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}=0\)

=> \(x+9=0\Rightarrow x=-9\) (không thỏa mãn)

Vậy không xác định được giá trị x

d

ĐK: x > 0 

\(A>5\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}>5\)

\(\Leftrightarrow x+9>5.2\sqrt{x}\Leftrightarrow x+9>10\sqrt{x}\)

\(\Leftrightarrow\left(x+9\right)^2>\left(10\sqrt{x}\right)^2=100x\)

<=> \(x^2+18x+81-100x>0\)

<=> \(x^2-82x+81>0\)

<=> \(x^2-81x-x+81>0\)

<=> \(x\left(x-81\right)-\left(x-81\right)>0\)

<=> \(\left(x-1\right)\left(x-81\right)>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1>0\\x-81>0\end{matrix}\right.\\\left[{}\begin{matrix}x-1< 0\\x-81< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x>81\end{matrix}\right.\\\left[{}\begin{matrix}x< 1\\x< 81\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>81\\x< 81\end{matrix}\right.\)

Vậy để A > 5 thì x > 81 và 0 < x < 81

a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì

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

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

c: M>0

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

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-2>0\)

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

=>x>4

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 \(x=\dfrac{66-16\sqrt{2}}{8649}\)

quy đồng lên là xong. Rút gọn nữa

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)