\(2,\\ 1,=20\sqrt{3}+20\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=40\sqrt{3}+\sqrt{3}=41\sqrt{3}\\ 2,A=\dfrac{2\sqrt{x}-9-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{2\sqrt{x}-x+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\left(4>0\right)\\ \Leftrightarrow x< 9\Leftrightarrow0\le x< 9\)

\(3,\\ 1,A=\sqrt{2}-1-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\sqrt{2}-1-\sqrt{2}=-1\\ 2,\\ a,P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\left(x\ge0;x\ne4\right)\\ P=\dfrac{4\left(\sqrt{x}+2\right)}{4\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ b,P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)

Bài 2: 

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3

Vậy: A ⋮ 3

_____________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7

Vậy: A ⋮ 7

___________________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5

Vậy: A ⋮ 5 

a) x * 5 + x * 3 = 720

           x*(5+3) = 720

                 x* 8= 720

                    x = 720: 8

                    x = 90

b) x : 9 = 1126(dư 7)

         x = 1126 * 9 + 7

          x = 10141

c)x * 6 + x + x + x = 954

           x * 6 + x * 3= 954

             x * (6 + 3) = 954

                      x * 9 = 954

                           x = 954 : 9

                            x= 106

mik chúc bạn học tốt, nhớ t.i.c.h cho mik nha^^

Bài 1:

a: \(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}=x\)

=>\(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)

=>\(x-\dfrac{1}{2}>=0\)

=>\(x>=\dfrac{1}{2}\)

b: \(\left|1-3x\right|+1=3x\)

=>\(\left|1-3x\right|=3x-1\)

=>\(1-3x< =0\)

=>3x-1>=0

=>3x>=1

=>\(x>=\dfrac{1}{3}\)

Bài 2:

a: \(C=\left|5-x\right|+x=\left|x-5\right|+x\)

TH1: x>=5

\(C=x-5+x=2x-5\)

TH2: x<5

C=5-x+x=5

b: D=|2x-1|-x

TH1: x>=1/2

\(D=2x-1-x=x-1\)

TH2: \(x< \dfrac{1}{2}\)

D=1-2x-x=1-3x

Gọi số cần lập là \(\overline{a_1a_2a_3a_4a_5a_6}\)

\(\Rightarrow a_1\ge3\)

\(\Rightarrow a_1\) có 4 cách chọn (3;4;5;6)

5 chữ số còn lại có \(A_5^5=120\) cách

\(\Rightarrow4.120=480\) số thỏa mãn