bài 1 a)

\(\frac{15}{7}\times\frac43+\frac63\)

=\(\frac{60}{21}+\frac63\)

=\(\frac{60}{21}+\frac{42}{21}\)

=\(\frac{102}{21}\)

=\(\frac{34}{7}\)

b)\(\frac{-13}{41}\times\frac{17}{11}\times\frac{11}{17}\)

=\(\frac{-13}{41}\times\left(\frac{17}{11}\times\frac{11}{17}\right)\)

=\(\frac{-13}{41}\times1\)

=\(\frac{-13}{41}\)

a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)

(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)

\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)

\(\frac{44}{7}.x=\frac{-77}{35}\)

x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)

(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1

(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)

(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)

\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)

(44\(x\) + 3) : \(\frac{11}{5}\) = - 4

44\(x\) + 3 = - 4 x \(\frac{11}{5}\)

44\(x\) + 3 = - \(\frac{44}{5}\)

44\(x\) = - \(\frac{44}{5}\) - 3

44\(x\) = - \(\frac{59}{5}\)

\(x=-\frac{59}{5}:44\)

\(x\) = - \(\frac{59}{250}\)

Vậy \(x\) = - \(\frac{59}{250}\)

a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)

\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)

\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)

=>-5/3x=55/212

hay x=-33/212

c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)

\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)

=>x+3=19

hay x=16

A = (19\(\frac58\) : \(\frac{7}{12}\) - 13\(\frac14\): \(\frac72\)) x \(\frac45\) (xem lại đề bài)

B = (9\(\frac{30303}{80808}\) + 7\(\frac{303030}{808080}\)) + 4,03

B = (9\(\frac38\) + 7\(\frac38\)) + 4,03

B =(9 + 7) + (\(\frac38+\frac38\)) + 4,03

B = 16 + 3/4 + 4,03

B = 16 + 0,75 + 4,03

B = 16,75 + 4,03

B = 20,78

a) x = 1

a.X=\(\frac{25}{7}\)

a. 32 = 2⁵ => n thuộc tập 1; 2; 3; 4

b. \(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{4}+\frac{2}{3}=\frac{11}{12}\)

\(\Rightarrow x=\frac{12}{11}\)

c. p nguyên tố => \(p\ge2\) => 5^2p luôn có dạng A25

=> 5^2p+2015 chẵn

=> 2014^2p + q³ chẵn

Mà 2014^2p chẵn => q³ chẵn => q chẵn => q = 2

=> 5^2p + 2015 = 2014^2p+8

=> 5^2p+2007 = 2014^2p

2014 có mũ dạng 2p => 2014^2p có dạng B6

=> 5^2p = B6 - 2007 = ...9 (vl)

(hihi câu này hơi sợ sai)

d. \(17A=\frac{17^{19}+17}{17^{19}+1}=1+\frac{16}{17^{19}+1}\), \(17B=\frac{17^{18}+17}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

\(17^{19}+1>17^{18}+1\Rightarrow\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)

de thi chon hoc sinh gioi nay

5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)

(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)

5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)

5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)

5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)

5 : x = \(\frac{1}{17}\)

x = 5 : \(\frac{1}{17}\)

x = 5 x 17

x = 85

Vậy x = 85

1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19

3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19

1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19

1- 1/(x+3) = 1 - 1/19

1/(x+3) = 1/19

x + 3 = 19

x = 19 - 3

x = 16

Vậy x = 16

1

x + 7/12= 15/18

x= 15/18-7/12

x= 1/4

k mk nhé

3

4/10= 2/5

\(\frac{x}{15}\)=\(\frac{2}{5}\)

x= 15:5x2= 6

x= 6

k mk nhé