Bài 1 

a, \(-\dfrac{6}{11}:\left(\dfrac{3}{5}.\dfrac{4}{11}\right)=-\dfrac{6}{11}:\dfrac{12}{55}=-\dfrac{6}{11}.\dfrac{55}{12}=-\dfrac{5}{2}\)

b, \(\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42+5-22}{72}=\dfrac{25}{72}\)

Bài 2 

a, \(x+\dfrac{1}{3}=\dfrac{7}{26}.\left(-\dfrac{13}{6}\right)\)

\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{7}{12}\Leftrightarrow x=-\dfrac{7}{12}-\dfrac{1}{3}=\dfrac{-7-4}{12}=-\dfrac{11}{12}\)

b, \(\dfrac{x}{150}=\dfrac{5}{6}.\left(-\dfrac{7}{25}\right)\)

\(\Leftrightarrow\dfrac{x}{150}=-\dfrac{7}{30}\Leftrightarrow x=\left(-\dfrac{7}{30}\right).150=-35\)

c, \(x.\dfrac{3}{5}=\dfrac{2}{5}\Leftrightarrow x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{5}.\dfrac{5}{3}=\dfrac{2}{3}\)

a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)

\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)

\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)

\(\dfrac{x}{3}=\dfrac{29}{15}\)

\(x=5,8\)

b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)

\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)

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

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

\(x=\dfrac{7}{4}\)

\(\left(2n+1\right)⋮\left(2n-3\right)\)

Vì \(\left(2n+1\right)⋮\left(2n-3\right)\Rightarrow\left(2n-3\right)+4⋮\left(2n-3\right)\)

Vì \(2n-3⋮2n-3\Rightarrow4⋮2n-3\)

Vì \(4⋮\left(2n-3\right)\Rightarrow2n-3\) là \(Ư\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)

Lập bảng:

2n + 1 ⋮ 2n - 3.

⇒ (2n - 3) + 4 ⋮ 2n - 3                     Do 2n - 3 ⋮ 2n - 3.

Ta cần 4 ⋮ 2n - 3 ⇒ 2n - 3 \(\in\) Ư(4)=\(\left\{1;2;4\right\}\).

Mà 2n - 3 là số lẻ ⇒ 2n - 3 = 1.

2n - 3 = 1 ⇒ n = 2.

Vậy n = 2.

3/1x3 + 3/3x5 + .... + 3/49x51

= 3 ( 1/1x3 + 1/3x5 + .... + 1/49x51 )

= 3/2 ( 2/1x3 + 2/3x5 +.... + 2/49x51 )

= 3/2 ( 1 - 1/3 + 1/3 - 1/5 +...+ 1/49 - 1/51 )

= 3/2 ( 1 - 1/51 ) = 3/2 x 50/51 = 25/17

giúp mình vs

Lời giải:

Để $A$ nhận giá trị nguyên thì $2x-3\vdots 2-3x$

$\Rightarrow 3(2x-3)\vdots 2-3x$

$\Rightarrow 6x-9\vdots 2-3x$

$\Rightarrow 2(3x-2)-5\vdots 2-3x$

$\Rightarrow 5\vdots 2-3x$

$\Rightarow 2-3x\in\left\{\pm 1; \pm 5\right\}$

$\Rightarrow x\in\left\{\frac{1}{3}; 1; -1; \frac{7}{3}\right\}$

Vì $x$ nguyên nên $x\in\left\{1; -1\right\}$

Thử lại thấy thỏa mãn.

            A =                 \(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{3}{2^2}+\dfrac{3}{2^3}+.....+\dfrac{3}{2^{2021}}+\dfrac{3}{2^{2022}}\)

     \(2\times\)A =             1 + 3+   \(\dfrac{3}{2}\) +\(\dfrac{3}{2^2}\)  + \(\dfrac{3}{2^3}\)+...........+\(\dfrac{3}{2^{2021}}\)

2 \(\times\) A - A =           4 - \(\dfrac{1}{2}\) - \(\dfrac{3}{2^{2022}}\)

             A =          \(\dfrac{7}{2}\)    - \(\dfrac{3}{2^{2022}}\)

            B =                  2 \(\times\dfrac{3}{2^{2023}}\)

      A - B  =         \(\dfrac{7}{2}-\dfrac{3}{2^{2022}}\)  - 2 \(\times\) \(\dfrac{3}{2^{2023}}\)

     A - B =           \(\dfrac{7}{2}\)   - \(\dfrac{3}{2^{2022}}\) - \(\dfrac{3}{2^{2022}}\)

    A - B =            \(\dfrac{7}{2}\) - \(\dfrac{6}{2^{2022}}\)

   A - B =            \(\dfrac{7}{2}\) - \(\dfrac{3}{2^{2021}}\)

\(8-\dfrac{4}{5}=\dfrac{5}{x^2}\\ \Leftrightarrow\dfrac{40-4}{5}=\dfrac{5}{x^2}\\ \Leftrightarrow36x^2-25\\ \Leftrightarrow x^2=\dfrac{25}{36}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)

\(8-\dfrac{4}{5}=\dfrac{5}{x^2}\)

\(\dfrac{5}{36}=\dfrac{5}{6^2}=\dfrac{5}{x^2}\)

Vậy 6 = x