a, \(\dfrac{4}{7}\). \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)

     \(\dfrac{4}{7}\).\(\dfrac{a}{b}\)        =   \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)

     \(\dfrac{4}{7}\).\(\dfrac{a}{b}\)        = \(\dfrac{8}{21}\)

          \(\dfrac{a}{b}\)       = \(\dfrac{8}{21}\):\(\dfrac{4}{7}\)

           \(\dfrac{a}{b}\)      = \(\dfrac{2}{3}\)

b, \(\dfrac{a}{b}\) + \(\dfrac{2}{3}\).\(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)

     \(\dfrac{a}{b}\) + \(\dfrac{2}{9}\) = \(\dfrac{2}{3}\)

      \(\dfrac{a}{b}\)        = \(\dfrac{2}{3}\) - \(\dfrac{2}{9}\)

       \(\dfrac{a}{b}\)       = \(\dfrac{4}{9}\)

c, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}.\)\(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)

    \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\)     = \(\dfrac{2}{7}\)

    \(\dfrac{a}{b}\)           = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)

    \(\dfrac{a}{b}\)          = \(\dfrac{13}{21}\)

d, \(\dfrac{11}{13}\): \(\dfrac{a}{b}\): \(\dfrac{2}{3}\) = 2\(\dfrac{7}{13}\)

     \(\dfrac{11}{13}\): \(\dfrac{a}{b}\):\(\dfrac{2}{3}\) = \(\dfrac{33}{13}\)

      \(\dfrac{11}{13}\): \(\dfrac{a}{b}\)    = \(\dfrac{33}{13}\) \(\times\) \(\dfrac{2}{3}\)

       \(\dfrac{11}{13}\): \(\dfrac{a}{b}\)   = \(\dfrac{66}{39}\)

                \(\dfrac{a}{b}\) = \(\dfrac{11}{13}\) : \(\dfrac{66}{39}\)

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

a, \(\dfrac{4}{7}\) \(\times\) \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)

      \(\dfrac{4}{7}\) \(\times\) \(\dfrac{a}{b}\)      = \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)

       \(\dfrac{4}{7}\) \(\times\) \(\dfrac{a}{b}\)     = \(\dfrac{8}{21}\)

                \(\dfrac{a}{b}\)    = \(\dfrac{8}{21}\): \(\dfrac{4}{7}\)

                  \(\dfrac{a}{b}\)    = \(\dfrac{2}{3}\)

b, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)

      \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\)        = \(\dfrac{2}{7}\)

       \(\dfrac{a}{b}\)             = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)

         \(\dfrac{a}{b}\)           = \(\dfrac{13}{21}\)

1) ta có: a(b^2 -1)(c^2 -1)+b(a^2 -1)(c^2 -1)+c(a^2-1)(b^2-1)

=(ab^2 -a)(c^2-1)+(ba^2 -b)(c^2-1)+(ca^2-c)(b^2-1)

 đén đây nhân bung ra hết rồi rút gọn và thay a+b+c=abc là đc

a+5 chia hết cho 11;13

=> a+5 thuộc BC(11;13) ; BCNN(11;13) = 143

=> a+5 = 143k=> a = 143k -5 ; với k thuộc N*

vì 99<a<1000=>99<143k-5<1000 =>0,72..<k< 7,02..

=>a nhỏ nhất ; khi k = 1

=>a =143 -5 = 138

Vậy a =138

Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=11\cdot\frac{13}{17}\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{143}{17}\)

\(\Rightarrow\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{a+c}{c+a}=\frac{143}{17}\)

\(\Rightarrow1+1+1+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=\frac{143}{17}\)

\(\Rightarrow A=\frac{143}{17}-3=\frac{92}{17}\)

a)Ta có:

 A= 1/1.2+1/2.3+1/3.4+.....+1/99.100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

b)Ta có:

B= 1/11+1/12+1/13+1/14+1/15+...+1/50

=(1/11+1/50)+(1/12+1/49)+...+(1/30+1/31)

=61/11.50+61/12.49+...+61/30.31

=61.(1/11.50+1/12.49+...+1/30.31)

Mình xin lỗi chỉ làm được đến đây vì dạng tính B mình không tốt lắm ◕◡◕

\(B=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)>\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)=> \(B>\frac{20}{30}+\frac{20}{50}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>1\)

mà \(A=\frac{99}{100}