\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))

\(=\left[\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)\(=\left[\dfrac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

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

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

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

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

\(\Rightarrow P=\dfrac{M}{N}+1=\dfrac{2}{\sqrt{x}+2}+1\)

Ta thấy: \(\sqrt{x}\ge0\forall x\)

\(\Rightarrow\sqrt{x}+2\ge2\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}+1\le2\forall x\)

\(\Rightarrow Max_P=2\Leftrightarrow\dfrac{2}{\sqrt{x}+2}+1=2\)

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

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

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

#Urushi☕

Bạn tự rút gọn nha .

c) Ta có : \(P\text{=}\dfrac{M}{N}+1\text{=}\dfrac{2}{\sqrt{x}+2}+1\)

Để P có giá trị lớn nhất.

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}cóGTLN\)

\(\Leftrightarrow\sqrt{x}+2cóGTNN\)

Mà : \(\sqrt{x}+2\ge2\)

\(\Rightarrow\) Để : \(\left(\sqrt{x}+2\right)_{min}\) \(\Leftrightarrow\sqrt{x}\text{=}0\Leftrightarrow x\text{=}0\)

Vậy............

\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

cau b nua

\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)

\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)

Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)

a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)

=>x+6<0

=>x<-6

b: ĐKXĐ: (-2)/(6-x)>=0

=>6-x<0

=>x>6

c: ĐKXĐ: (-x+3)/(-6)>=0

=>-x+3<=0

=>-x<=-3

=>x>=3

d: ĐKXĐ: (7x-1)/-9>=0

=>7x-1<=0

=>x<=1/7

e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0

=>x+2>=0

=>x>=-1

f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0

=>x-2>=0

=>x>=2

câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm

1) So sánh:

N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)

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

\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)

M = \(\sqrt{18}-\sqrt{8}\)

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

\(=\sqrt{2}\)

Ta có: \(1=\sqrt{1}\)

Mà 1 < 2

\(\Rightarrow\sqrt{1}< \sqrt{2}\)

Hay 1 \(< \sqrt{2}\)

Vậy N < M
 

\(A=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\left(x>0,x\ne1\right)\)

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

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

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

\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4;9\right)\)

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

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

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

\(C=\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

b, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

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

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

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

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

b) Ta có: \(B=\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)

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

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

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

\(=\dfrac{\sqrt{x}-3-5+\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

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

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

a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

\(=\dfrac{x-4\sqrt{x}+4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-5\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

1: ĐKXĐ: \(a>-2\)

2: ĐKXĐ: \(x\ne2\)

3: ĐKXĐ: \(a\in\varnothing\)

1)
\(-\dfrac{1}{\sqrt{a+2}}\) có nghĩa khi \(\sqrt{a+2}>0\)
=>a+2>0
    a>-2
2)
\(\sqrt{\dfrac{3}{\left(x-2\right)^2}}=\dfrac{\sqrt{3}}{\sqrt{\left(x-2\right)^2}}\) 
mà \(\left(x-2\right)^2>0=>\sqrt{\left(x-2\right)^2}>0vớimọix\)
3)
\(\sqrt{\dfrac{-3}{a^2-4a+4}}=\sqrt{\dfrac{-3}{\left(a-2\right)^2}}cónghĩakhi\left(a-2\right)^2< 0mà\left(a-2\right)^2>0=>biểuthứckocónghĩavớimọia\)
 

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