\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)

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

b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Kết hợp đk

\(\Rightarrow x\in\left\{4;16;64\right\}\)

giai giúp mình với

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

\(\Leftrightarrow6\sqrt{x}+9⋮3\sqrt{x}-1\)

\(\Leftrightarrow3\sqrt{x}-1\in\left\{-1;1;11\right\}\)

\(\Leftrightarrow3\sqrt{x}\in\left\{0;12\right\}\)

hay \(x\in\left\{0;16\right\}\)

a) ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp đk:

\(\Leftrightarrow x\in\left\{0\right\}\)

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

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

Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)

\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(minP=-1\Leftrightarrow x=0\)

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)

\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)

Dấu \("="\Leftrightarrow x=0\)

a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)

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

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

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

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

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

b) Với \(x\ge0;x\ne1;x\ne4\):

Thay \(x=3+2\sqrt{2}\) vào \(P\), ta được:

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

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)

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

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

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

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

c) Với \(x\ge0;x\ne1;x\ne4\),

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

Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên

\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)

\(\Rightarrow\sqrt{x}-1\in\left\{1;3;-1;-3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)

\(\Rightarrow x\in\left\{4;16;0\right\}\)

Kết hợp với ĐKXĐ của \(x\), ta được:

\(x\in\left\{0;16\right\}\)

Vậy: ...

\(\text{#}Toru\)

Bài 2 :

a, Ta có : \(x^2-5x+4< 0\)

\(\Leftrightarrow x^2-x-4x+4< 0\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)< 0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)< 0\)

Vậy ...

b, Ta có : \(\dfrac{x-3}{x+1}< 1\)

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

\(\Leftrightarrow\dfrac{x-3-x-1}{x+1}=\dfrac{-4}{x+1}< 0\)

Thấy - 4 < 0

Nên để \(-\dfrac{4}{x+1}< 0\) <=> x + 1 > 0 ( TH A, B trái dấu )

Vậy ...

Bài 1: 

a) ĐKXĐ: \(x\ge2\)

Ta có: \(3\sqrt{x-2}+\sqrt{25x-50}=2^5\)

\(\Leftrightarrow3\sqrt{x-2}+5\sqrt{x-2}=32\)

\(\Leftrightarrow8\sqrt{x-2}=32\)

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

\(\Leftrightarrow x-2=16\)

hay x=18(thỏa ĐK)

Vậy: S={18}

\(a,ĐK:x>0;x\ne4\\ E=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}-2}{2\sqrt{x}}\\ b,x=19-8\sqrt{3}=\left(4-\sqrt{3}\right)^2\\ \Leftrightarrow E=\dfrac{4-\sqrt{3}-2}{2\left(4-\sqrt{3}\right)}=\dfrac{\left(2-\sqrt{3}\right)\left(4+\sqrt{3}\right)}{26}=\dfrac{5-2\sqrt{3}}{26}\\ c,E=-1\Leftrightarrow\sqrt{x}-2=-2\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{4}{9}\left(tm\right)\\ d,E=\dfrac{1}{\sqrt{x}}\Leftrightarrow\dfrac{\sqrt{x}-2}{2}=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)

\(e,E>0\Leftrightarrow\sqrt{x}-2>0\left(2\sqrt{x}>0\right)\Leftrightarrow x>4\\ f,E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}=\dfrac{1}{2}-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\left(-\dfrac{1}{\sqrt{x}}< 0\right)\\ g,\dfrac{1}{E}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(4\right)=\left\{-1;0;1;2;4\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;2;3;4;6\right\}\\ \Leftrightarrow x\in\left\{1;9;16;36\right\}\left(x\ne4\right)\\ h,x>4\Leftrightarrow\sqrt{x}-2>0\\ \Leftrightarrow E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}>0\Leftrightarrow E\ge\sqrt{E}\)

a: \(A=5\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-1\)

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

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

b: A=B

=>căn x=-căn x+1

=>căn x=1/2

=>x=1/4