3.

\(F=\dfrac{k.\left|q_1.q_2\right|}{r^2}=\dfrac{9.10^9.\left|9.10^{-18}\right|}{0,1^2}=8,1.10^{-6}N\)

2. C

1.

\(F=\dfrac{k.\left|q_1.q_2\right|}{r^2}=\dfrac{9.10^9.\left|5.10^8.\left(-1,6.10^{-19}\right).5.10^8.\left(-1,6.10^{-19}\right)\right|}{0,02^2}=1,44.10^{-7}N\)

Bài 2:

a: \(=\dfrac{3}{5}\cdot\dfrac{6}{7}=\dfrac{18}{35}\)

b: =3/5+8/25

=15/25+8/25

=23/25

Bài 3:

a: =>x=2/3*2/5=4/25

b: =>x=4/9*2/3=8/27

Bài 1:
\(130050:452=287\)(dư 326)
\(19183:78=245\)(dư 73)
\(204\times1942=396168\)
Bài 2:
\(\dfrac{4}{9}+\dfrac{3}{7}=\dfrac{28}{63}+\dfrac{27}{63}=\dfrac{55}{63}\)
\(\dfrac{7}{15}-\dfrac{11}{30}=\dfrac{14}{30}-\dfrac{11}{30}=\dfrac{3}{30}=\dfrac{1}{10}\)

Q=m.c.(t2-t1)

<=>13,68.1000=0,3.380.(t2-t1)

<=>t2-t1=120 

=>t2=120+t1=120+20=140(độ C)

Nhiệt độ cuối của miếng đồng là 140 độ C

Độ tăng nhiệt độ của miếng chì :

\(\Delta t=\dfrac{Q}{m\cdot c}=\dfrac{4680}{0.3\cdot130}=120^0C\)

3: \(=\dfrac{\left(3-\sqrt{5}\right)^2}{\sqrt{5}-3}=\sqrt{5}-3\)

4: \(=\dfrac{\left(\sqrt{5}-2\right)^2}{\sqrt{5}-2}=\sqrt{5}-2\)

5: \(=\dfrac{8-2\sqrt{15}+4\sqrt{15}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{8+2\sqrt{15}}{\sqrt{5}+\sqrt{3}}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\sqrt{5}+\sqrt{3}}=\sqrt{5}+\sqrt{3}\)

6:

\(=\dfrac{8\sqrt{6}-11-4\sqrt{6}}{4\sqrt{2}-2\sqrt{3}}\)

\(=\dfrac{4\sqrt{6}-11}{4\sqrt{2}-2\sqrt{3}}=\dfrac{\sqrt{3}-2\sqrt{2}}{2}\)

1) \(\dfrac{3\sqrt{5}-5\sqrt{3}}{3-\sqrt{15}}=\dfrac{\sqrt{15}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)}=\sqrt{15}\)

2) \(\dfrac{5\sqrt{6}-6\sqrt{5}}{2\sqrt{15}-5\sqrt{2}}=\dfrac{\sqrt{30}\left(\sqrt{5}-\sqrt{6}\right)}{\sqrt{10}\left(\sqrt{6}-\sqrt{5}\right)}=-\sqrt{3}\)

3) \(\dfrac{14-6\sqrt{5}}{\sqrt{5}-3}=\dfrac{\left(3-\sqrt{5}\right)^2}{\sqrt{5}-3}=\sqrt{5}-3\)

4) \(\dfrac{9-4\sqrt{5}}{\sqrt{5}-2}=\dfrac{\left(2-\sqrt{5}\right)^2}{\sqrt{5}-2}=\sqrt{5}-2\)

5) \(\dfrac{\left(\sqrt{3}-\sqrt{5}\right)^2+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}=1\)

6) \(\dfrac{8\sqrt{6}-\left(\sqrt{3}+2\sqrt{2}\right)^2}{4\sqrt{2}-2\sqrt{3}}=\dfrac{8\sqrt{6}-11-4\sqrt{6}}{4\sqrt{2}-2\sqrt{3}}\)

\(=\dfrac{4\sqrt{6}-11}{2\left(2\sqrt{2}-\sqrt{3}\right)}=\dfrac{(\sqrt{3}-2\sqrt{2})^2}{2\left(\sqrt{3}-2\sqrt{2}\right)}=\dfrac{\sqrt{3}-2\sqrt{2}}{2}\)

câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\)  (đk \(x\)≠ -1; 1)

          \(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)

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

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

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

b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)

    2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\)  - 2)

   \(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)

c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\)  đk \(x\) ≠ 2; -2

\(\dfrac{x}{2x-4}\)  =   \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\) 

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

\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\)  = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)

\(\dfrac{4x^2-3x+5}{x^3-1}\) =  \(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) Đk \(x\) ≠ 1
\(\dfrac{6}{x-1}\) = \(\dfrac{6.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

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