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\(1.\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)=8\) \(2.a,b.A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\) ( x # 0 ; x # -1 ; x # 1 )
\(A=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{x^2-1}.\dfrac{x+2003}{x}\)
\(A=\dfrac{x^2-1}{x^2-1}.\dfrac{x+2003}{x}=\dfrac{x+2003}{x}\)
c. \(A=1+\dfrac{2003}{x}\)
Để A ∈ Z ⇒ x ∈ { 1 ; -1 ; 2003 ; - 2003 )
KL...............
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
\(a.\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\dfrac{1}{x-1}\right)=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\dfrac{x-2}{x-1}=\dfrac{\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|}{\left|x-2\right|}.\dfrac{x-2}{x-1}\left(x>1\right)\)
Tới đây dễ r , bạn tự chia TH ra làm nhé :D
\(b.\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{\sqrt{x^3}-x}{1-\sqrt{x}}=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}+\dfrac{x\sqrt{x}-x}{\sqrt{x}-1}=-2\sqrt{x-1}+x\left(x\ge1\right)\)
Bạn ơi câu a có vẻ có vấn đề ý. Nếu bạn áp dụng HĐT thì phải là√(x-2)2 chứ nhỉ. Mong bạn giải đáp
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
a) P = \(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) + \(\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
P = \(\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\) - \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) - \(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
P = \(\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) = \(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Để \(\sqrt{P}\) có nghĩa P ≥ 0 ⇒ \(\sqrt{x}-1\) > 0 ⇒ x = 1
P = \(1+\dfrac{2}{\sqrt{x}-1}>1\)
Xét \(P-\sqrt{P}\) = \(\sqrt{P}\left(\sqrt{P}-1\right)\)
Mà \(\sqrt{P}>0\)
Vì P > 1 ⇒ \(\sqrt{P}>\sqrt{1}\Rightarrow\sqrt{P}>1\Rightarrow\sqrt{P}-1>0\Rightarrow P-\sqrt{P}>0\Leftrightarrow P>\sqrt{P}\)
c) Tìm x để \(\dfrac{1}{P}\in Z\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\Rightarrow\dfrac{1}{P}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
\(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le\dfrac{2}{1}\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge-1\)
\(\Rightarrow-1\le\dfrac{1}{P}< 1\Rightarrow\dfrac{1}{P}\in\left\{-1;0\right\}\)
\(với\dfrac{1}{P}=-1\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}=0\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
\(với\dfrac{1}{P}=0\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=0\)
\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(loại\right)\)
Vậy x=0 thì \(\dfrac{1}{P}\in Z\)
CHÚC BẠN HỌC TỐT 
1. a) \(A=\left(\dfrac{\sqrt{x}-1+x-\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}\right).\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)ĐK x\(\ne\)0,1
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(x-1\right)}=\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
b) A<-1 <=> \(\dfrac{2\sqrt{x}}{x-\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}+1< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x-\sqrt{x}}{x-\sqrt{x}}< 0\)\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x-\sqrt{x}}< 0\)
\(\Leftrightarrow x-\sqrt{x}< 0\) (vì \(x+\sqrt{x}>0\left(\forall x>0\right)\))
\(\Leftrightarrow x< \sqrt{x}\Leftrightarrow x^2< x\Leftrightarrow x^2-x< 0\)
\(\Leftrightarrow x\in\left(0;1\right)\Leftrightarrow0< x< 1\)
a) ĐKXĐ: x≠0;x≠1;x>0
b) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-x}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{x-1}{x+\sqrt{x}+1}=\left(\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)}{x\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\left(\dfrac{2+\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\)