Bài 6:

\(a)P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{33\cdot37}+\dfrac{2}{37\cdot41}\\ =\dfrac{1}{2}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{33\cdot37}+\dfrac{4}{37\cdot41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\dfrac{40}{41}\\ =\dfrac{20}{41}\\ b)Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\\ =\dfrac{6}{7}\cdot\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\dfrac{119}{242}\\ =\dfrac{51}{121}\)

Bài 5:

a: Để A>0 thì \(\dfrac{2a-1}{-5}>0\)

=>2a-1<0

=>\(a< \dfrac{1}{2}\)

b: Để A<0 thì \(\dfrac{2a-1}{-5}< 0\)

=>2a-1>0

=>2a>1

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

c: Để A=0 thì \(\dfrac{2a-1}{-5}=0\)

=>2a-1=0

=>2a=1

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

Bài 6:

a: \(P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{37\cdot41}\)

\(=\dfrac{2}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{37\cdot41}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{41}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{41}\right)=\dfrac{1}{2}\cdot\dfrac{40}{41}=\dfrac{20}{41}\)

b: \(Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\)

\(=\dfrac{6}{7}\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{121}\right)=\dfrac{6}{7}\cdot\dfrac{119}{242}=\dfrac{51}{121}\)

Bài  1d;

(1\(\dfrac{4}{7}\) - 2\(\dfrac{2}{5}\)):(1\(\dfrac{1}{8}\) - 4\(\dfrac{3}{4}\))

= (\(\dfrac{11}{7}\) - \(\dfrac{12}{5}\)):(\(\dfrac{9}{8}\) - \(\dfrac{19}{4}\))

= (\(\dfrac{55}{35}\) - \(\dfrac{84}{35}\)): (\(\dfrac{9}{8}\) - \(\dfrac{38}{8}\))

= - \(\dfrac{29}{35}\): (- \(\dfrac{29}{8}\))

= - \(\dfrac{29}{35}\) x \(\dfrac{8}{-29}\)

= \(\dfrac{8}{35}\) 

Bài 2:

a; - \(\dfrac{2}{11}\) + \(\dfrac{6}{7}\) + \(\dfrac{1}{2}\) + \(\dfrac{-9}{11}\) + \(\dfrac{1}{7}\)

=  - (\(\dfrac{2}{11}\) + \(\dfrac{9}{11}\)) + (\(\dfrac{6}{7}\) + \(\dfrac{1}{7}\)) + \(\dfrac{1}{2}\)

= - \(\dfrac{11}{11}\) + \(\dfrac{7}{7}\) + \(\dfrac{1}{2}\)

= -1 + 1 + \(\dfrac{1}{2}\)

= 0 + \(\dfrac{1}{2}\)

= \(\dfrac{1}{2}\) 

Bài 1:

a;  \(\dfrac{4}{5}\) - (-\(\dfrac{2}{7}\)) + ( - \(\dfrac{7}{10}\))

=   \(\dfrac{4}{5}\) + \(\dfrac{2}{7}\) - \(\dfrac{7}{10}\)

=    \(\dfrac{56}{70}\) + \(\dfrac{2\times10}{7\times10}\) - \(\dfrac{7}{10}\)

=     \(\dfrac{56}{70}\) + \(\dfrac{20}{70}\) - \(\dfrac{49}{10}\)

= \(\dfrac{76}{70}\) - \(\dfrac{49}{10}\)

= \(\dfrac{27}{10}\) 

b; \(\dfrac{2}{15}\).\(\dfrac{5}{8}\) - \(\dfrac{5}{6}\).(-\(\dfrac{2}{3}\))

= \(\dfrac{1}{12}\) + \(\dfrac{5}{9}\)

= \(\dfrac{3}{36}\) + \(\dfrac{20}{36}\)

= \(\dfrac{23}{36}\) 

Bài 1c;

(5\(\dfrac{2}{5}\).7\(\dfrac{7}{12}\)).(13\(\dfrac{8}{9}\).7\(\dfrac{5}{13}\))

= (\(\dfrac{27}{5}\).\(\dfrac{91}{12}\)).(\(\dfrac{125}{9}\).\(\dfrac{96}{13}\))

= \(\dfrac{819}{20}\).\(\dfrac{4000}{39}\)

= 4200

Bài 2b;

(\(\dfrac{13}{5}\) + \(\dfrac{7}{16}\)) - (\(\dfrac{15}{16}\) - \(\dfrac{6}{15}\))

= \(\dfrac{13}{5}\) + \(\dfrac{7}{16}\) - \(\dfrac{15}{16}\) + \(\dfrac{6}{15}\)

= (\(\dfrac{7}{16}\) - \(\dfrac{15}{16}\)) + (\(\dfrac{13}{5}\) +\(\dfrac{6}{15}\))

= - \(\dfrac{1}{2}\) + (\(\dfrac{39}{15}\) + \(\dfrac{6}{15}\))

= - \(\dfrac{1}{2}\) +  3

= - \(\dfrac{1}{2}\) + \(\dfrac{6}{2}\)

=   \(\dfrac{5}{2}\) 

Bài 2c;

  (\(-\dfrac{5}{7}\)).\(\dfrac{2}{11}\) + (-\(\dfrac{5}{7}\)).\(\dfrac{4}{11}\) + \(\dfrac{5}{7}\).-\(\dfrac{5}{11}\)

= - \(\dfrac{5}{7}\).(\(\dfrac{2}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{5}{11}\))

= - \(\dfrac{5}{7}\) .(\(\dfrac{6}{11}\) + \(\dfrac{5}{11}\))

= - \(\dfrac{5}{7}\).1

= - \(\dfrac{5}{7}\) 

    \(\dfrac{75}{26}\) : \(\dfrac{25}{39}\) : \(\dfrac{42}{55}\).\(\dfrac{21}{22}\)

=  \(\dfrac{75}{26}\) x \(\dfrac{39}{25}\) : \(\dfrac{42}{55}\) x \(\dfrac{21}{22}\)

= \(\dfrac{9}{2}\) : \(\dfrac{42}{55}\) x \(\dfrac{21}{22}\)

=  \(\dfrac{9}{2}\) x \(\dfrac{55}{42}\) x \(\dfrac{21}{22}\)

= \(\dfrac{9}{2}\) x (\(\dfrac{55}{42}\) x \(\dfrac{21}{22}\))

= \(\dfrac{9}{2}\) x \(\dfrac{5}{4}\)

= \(\dfrac{45}{8}\)

Bài 3:

a;  \(x\) + \(\dfrac{1}{5}\) = - \(\dfrac{4}{9}\)

     \(x\)          = - \(\dfrac{4}{9}\) - \(\dfrac{1}{5}\)

     \(x\)         = - \(\dfrac{20}{45}\) - \(\dfrac{9}{45}\)

    \(x\)          = - \(\dfrac{29}{45}\)

     Vậy  \(x=-\dfrac{29}{45}\)

Bài 3b;

2 - (\(x\) + \(\dfrac{3}{7}\)) = \(\dfrac{9}{-21}\)

     \(x+\dfrac{3}{7}\) = 2 + \(\dfrac{3}{7}\)

    \(x\)   =  2 + \(\dfrac{3}{7}\) - \(\dfrac{3}{7}\)

    \(x\)   = 2+ (\(\dfrac{3}{7}\) - \(\dfrac{3}{7}\))

    \(x\) =  2 

Vậy \(x=2\)

c; - \(\dfrac{2}{5}\) + 2.(\(\dfrac{3}{2}\) - \(x\)) = \(\dfrac{-7}{6}\)

             2.(\(\dfrac{3}{2}\) - \(x\)) = - \(\dfrac{7}{6}\) + \(\dfrac{2}{5}\)

           2.(\(\dfrac{3}{2}\) - \(x\))   =  - \(\dfrac{23}{30}\)

                 \(\dfrac{3}{2}\) - \(x\) = - \(\dfrac{23}{30}\) : 2

                \(\dfrac{3}{2}\) - \(x\) = - \(\dfrac{23}{60}\)

                \(x\)  = \(\dfrac{3}{2}\) + \(\dfrac{23}{60}\)

                \(x=\) \(\dfrac{113}{60}\)

Vậy \(x\) = \(\dfrac{113}{60}\) 

Bài 3d; 5 : (\(x\) - 1,2) + 1\(\dfrac{2}{3}\) = \(\dfrac{-5}{2}\)

           5 : (\(x\) - 1,2) + \(\dfrac{5}{3}\) =  \(\dfrac{-5}{2}\)

          5 : (\(x\) - 1,2) = \(\dfrac{-5}{2}\) - \(\dfrac{5}{3}\)

          5 : (\(x\) - 1,2) = - \(\dfrac{25}{6}\)

               \(x\) - 1,2 = 5 : - \(\dfrac{25}{6}\) 

             \(x\) - 1,2 =   - \(\dfrac{6}{5}\)

             \(x\)         = - \(\dfrac{6}{5}\) + 1,2

             \(x\)         = 0

Vậy \(x=0\) 

a; Phân số thoả mãn đề bài có dạng: \(\dfrac{x}{150}\) ; \(x\) \(\in\) Z

Theo bài ra ta có: \(\dfrac{7}{15}\) < \(\dfrac{x}{150}\) < \(\dfrac{12}{25}\)

       ⇒ \(\dfrac{7}{15}\) \(\times\) 150 < \(x\) < \(\dfrac{12}{25}\) \(\times\) 150

      ⇒ 70 < \(x\) < 72 vì \(x\) \(\in\) Z nên \(x\) = 71

Vậy phân số thoả mãn đề bài là: \(\dfrac{71}{150}\)

Bài 4 phân số thoả mãn đề bài có dạng: \(\dfrac{7}{x}\); (\(x\) \(\ne\) 0; \(x\in\) Z)

Theo bài ra ta có:   \(\dfrac{13}{-5}\) < \(\dfrac{7}{x}\) < \(\dfrac{-9}{4}\)

                           \(\dfrac{-819}{315}\) < \(\dfrac{-819}{-117x}\) < \(\dfrac{-819}{364}\)

                             315 < - 117\(x\) < 364 

                            \(\dfrac{315}{-117}\) > \(x\) > \(\dfrac{364}{-117}\)

                           -2\(\dfrac{81}{117}\) > \(x\) > - 3\(\dfrac{13}{117}\)

   Vì  \(x\) \(\in\) Z nên \(x=\) - 3

Vậy phân số thoả mãn đề bài là: \(-\dfrac{7}{3}\) 

            Bài 4c;

Viết ba số hữu tỉ xen giữa hai số hữu tỉ - \(\dfrac{3}{5}\) và  \(\dfrac{5}{-8}\)

       - \(\dfrac{3}{5}\) =  \(\dfrac{-15}{25}\) ; \(\dfrac{5}{-8}\) = \(\dfrac{-15}{24}\)

      \(\dfrac{-15}{25}\) = \(\dfrac{-60}{100}\);  \(\dfrac{-15}{24}\) = \(\dfrac{-60}{96}\)

Vậy ba phân số hữu tỉ nằm giữa hai số hữu tỉ - \(\dfrac{3}{5}\) và \(\dfrac{5}{-8}\) là:

       \(\dfrac{-60}{99}\); \(\dfrac{-60}{98}\); \(\dfrac{-60}{97}\) 

     Bài 7:

a; \(\dfrac{1}{x}\) - \(\dfrac{y}{6}\) = \(\dfrac{1}{3}\) (đk \(x\ne\) 0)

⇒    \(\dfrac{y}{6}\) = \(\dfrac{1}{x}\) - \(\dfrac{1}{3}\)

    \(\dfrac{y}{6}\) = \(\dfrac{3-x}{3x}\)

  3\(xy\) = 18 - 6\(x\)

  3\(xy\) + 6\(x\) = 18 

    \(3x\).(y + 2) = 18

     \(x\) = \(\dfrac{18}{3\left(y+2\right)}\) = \(\dfrac{6}{y+2}\)

     \(x\) \(\in\) z ⇔ 6 ⋮ y + 2

    y + 2 \(\in\) {- 6; - 3; - 2; -1; 1; 2; 3; 6}

    lập  bảng ta có:

Theo bảng trên ta có các cặp \(x;y\) nguyên thoả mãn đề bài là:

(\(x;y\)) = (-1; -8); (-2; -5); (-3; -4); (-6; -3); (6; -1); (3; 0); (2; 1); (1; 4)

       Bài 7

a; \(\dfrac{x}{2}\) + \(\dfrac{3}{y}\) = \(\dfrac{5}{4}\)

    \(\dfrac{xy+6}{2y}\) = \(\dfrac{5}{4}\) (y ≠ 0)

      4(\(xy\) + 6) = 10y

       4\(xy\) + 24 = 10y

    10y - 4\(xy\) = 24

      2y(\(5-2x\)) = 24

     y = \(\dfrac{24}{2\left(5-2x\right)}\) = \(\dfrac{12}{5-2x}\)  (\(x\in\) Z)

    y \(\in\) Z ⇔ 12 ⋮ \(5-2x\) 

⇒ \(5-2x\)  \(\in\) {-12; - 6; - 4; - 3; -2; -1; 1; 2; 3; 4; 6; 12}

Lập bảng ta có: 

Theo bảng trên ta có các cặp (\(x;y\)) nguyên thoả mãn đề bài là:

(\(x;y\)) = (4; - 4); (3; -12); (2; 12); (1; 4)