a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

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

\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)

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

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x\ge0\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)

a) 1308

b) 27

c) 350

d) 135

e) 285

f) 6

a) 1308

b) 27

c) 350

d) 135

e) 285

f) 6

1) Ư(5)={1; -1; 5; -5};     

 Ư(-30)={\(\pm1;\pm2;\pm3;\pm5;\pm6;\pm10;\pm15;\pm30\)}

Ư(19)={\(\pm1;\pm19\)}

Ư(22)=\(\left\{\pm1;\pm2;\pm11;\pm22\right\}\)

2b)BCNN(-8; -4)={8}

a)B(-8)={\(\pm8;\pm16;\pm24;\pm32;\pm40;...\)}

B(-4)\(\left\{\pm4;\pm8;\pm12;\pm16;\pm20;...\right\}\)

a: ĐKXĐ: x>=0; x<>1

\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: Khi \(x=4-2\sqrt3\) thì \(A=\frac{-5\cdot\sqrt{4-2\sqrt3}+2}{\sqrt{4-2\sqrt3}+3}=\frac{-5\left(\sqrt3-1\right)+2}{\sqrt3-1+3}\)

\(=\frac{-5\sqrt3+5+2}{\sqrt3+2}=\frac{-5\sqrt3+7}{\sqrt3+2}=\left(-5\sqrt3+7\right)\left(2-\sqrt3\right)\)

\(=-10\sqrt3+15+14-7\sqrt3=-17\sqrt3+29\)

c: \(A=\frac12\)

=>\(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac12\)

=>\(-10\sqrt{x}+4=\sqrt{x}+3\)

=>\(-11\sqrt{x}=-1\)

=>\(\sqrt{x}=\frac{1}{11}\)

=>x=1/121(nhận)

e: \(A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{-5\sqrt{x}+2+5\sqrt{x}+15}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}>0\forall x\) thỏa mãn ĐKXĐ

=>A>-5∀x thỏa mãn ĐKXĐ

a) ƯCLN của 28 ; 36 ; 35 là : 1

b) ƯCLN của 60 ; 120 ; 30 là : 30

a) 28=2².7

     36=2².3²

    35=5.7

ƯCLN(28,37,35)=1

ƯC(28,37,35)=Ư(1)=1

b)

60=2².3.5

120=2³.3.5

30=2.3.5 

ƯCLN(60,120,30)=2.3.5=30

ƯC(60,120,30)=Ư(30)=(1,2,3,5,6,10,15,40)

a, Gọi \(I\left(x;y\right)\) là tâm đường tròn ngoại tiếp \(\Delta ABC\)

\(\Rightarrow\left\{{}\begin{matrix}IA=IB\\IA=IC\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}IA^2=IB^2\\IA^2=IC^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(-3-x\right)^2+\left(6-y\right)^2=\left(1-x\right)^2+\left(-2-y\right)^2\\\left(-3-x\right)^2+\left(6-y\right)^2=\left(6-x\right)^2+\left(3-y\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=-5\\3x-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

Còn phần b,c,d,e nx bn C:

1. x=0

2. a) x = 1 và 1,1

b) x= 64,98; 64,99; 65;...

1.0

2.

a. 1  b.65

ta lấy : a,b > 0 ta có a,b > 0 ta làm a.b > 0 sẽ bằng 0 - 2 = âm 2 [ a,b] =240 và 16 ta lấy 240 - 16 + - 2 = 222

ta có : 240 -16 = 224 = 224 + 222 = 446 

nguyenhuyen   

nguyenhuyen

a: Ư(145)={1;5;29;145}

b: Ư(200)={1;2;4;5;8;10;20;25;40;50;100;200}