Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)

         = - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)

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

          = 2

2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8

   = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8

   =  \(\dfrac{2}{7}\) x 28 - 8

   = 8 - 8

    = 0 

\(7,M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right)\div\dfrac{2023}{2024}\\ M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\div\dfrac{2023}{2024}\\ M=\left(\dfrac{2\times\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\times\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\right)\div\dfrac{2023}{2024}\)

\(M=\left(\dfrac{2}{7}-\dfrac{1}{\dfrac{7}{2}}\right)\div\dfrac{2023}{2024}=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\div\dfrac{2023}{2024}\\ M=0\div\dfrac{2023}{2024}=0\)

𝓥𝓪̣̂𝔂 𝓜=0

\(8,A=\left(\dfrac{3}{1.8}+\dfrac{3}{8.15}+...+\dfrac{3}{106.113}\right)-\left(\dfrac{25}{50.55}+\dfrac{25}{55.60}+...+\dfrac{25}{95.100}\right)\\ A=\dfrac{3}{7}.\left(\dfrac{7}{1.8}+\dfrac{7}{8.15}+...+\dfrac{7}{106.113}\right)-5.\left(\dfrac{5}{50.55}+\dfrac{5}{55.60}+...+\dfrac{5}{95.100}\right)\\ A=\dfrac{3}{7}.\left(1-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{15}+...+\dfrac{1}{106}-\dfrac{1}{113}\right)-5.\left(\dfrac{1}{50}-\dfrac{1}{55}+\dfrac{1}{55}-\dfrac{1}{60}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\\ A=\dfrac{3}{7}.\left(1-\dfrac{1}{113}\right)-5.\left(\dfrac{1}{50}-\dfrac{1}{100}\right)\)

\(A=\dfrac{3}{7}.\dfrac{112}{113}-5.\dfrac{1}{100}=\dfrac{48}{113}-\dfrac{1}{20}\\ A=\dfrac{847}{2260}\)

a, Trên cùng một nửa mặt phẳng có bờ là tia ox vì \(\widehat{xOy}\) > \(\widehat{xOz}\) nên Oz nằm giữa hai tia Oy và Ox.

b, \(\widehat{xOy}\) = \(\widehat{xOz}\) +  \(\widehat{zOy}\) ⇒ \(\widehat{zOy}\)  = 80⁰ -  40⁰ = 40⁰

c, Oz nằm giữa hai tia Oy và Ox và \(\widehat{zOy}\) = \(\widehat{xOz}\)  nên OZ là tia phân giác của góc xOy

    A  =       \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{99}}\) + \(\dfrac{1}{3^{100}}\)

   3A  = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{99}}\)

3A - A = 1 - \(\dfrac{1}{3^{100}}\)

  2A     =  1 - \(\dfrac{1}{3^{100}}\)

     A     = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{100}}\) < \(\dfrac{1}{2}\) 

anh ơi bằng -5/7 anh nhé

2x + 3/4 = 1/2 : -5/4 - 1/2 x

2x + 1/2 x = -2/5 - 3/4

5/2 x = -23/20

x = -23/20 : 5/2 = -23/50

Vậy x = -23/50

19/57 . 2 4/7 - 19/57 . 1 5/7 - 19/57

= 19/57 x  ( 2 4/7 - 1 5/7 )

= 19/57 x 6/7

= 2/7

\(\dfrac{19}{57}.2\dfrac{4}{7}-\dfrac{19}{57}.1\dfrac{5}{7}-\dfrac{19}{57}\)

\(\text{=}\dfrac{19}{57}.\left(2\dfrac{4}{7}-1\dfrac{5}{7}-1\right)\)

\(\text{=}\dfrac{19}{57}.\left(\dfrac{-1}{7}\right)\)

\(\text{=}\dfrac{-1}{21}\)

c) \(2021,2345.2020,1234+2021,2345.\left(-2020,1234\right)\)

\(\text{=}2021,2345.\left(2020,1234-2020,1234\right)\)

\(\text{=}2021,2345.0\)

\(\text{=}0\)

d)\(4,75+\left(\dfrac{-1}{2}\right)^3+0,5^2-3.\dfrac{-3}{8}\)

\(\text{=}4,75-\dfrac{1}{8}+\dfrac{1}{4}+\dfrac{9}{8}\)

\(\text{=}\left(4,75+0,25\right)+\left(\dfrac{9}{8}-\dfrac{1}{8}\right)\)

\(\text{=}1+1\)

\(\text{=}2\)

Mình nhầm ở chỗ phần d nha.

Kết quả cuối cùng phải là :

\(\left(4,75+0,25\right)+\left(\dfrac{9}{8}-\dfrac{1}{8}\right)\)

\(\text{=}5+1\)

\(\text{=}6\)

a) \(12,4\cdot6\dfrac{1}{4}+\left(-12,4\right)\cdot\left(-2,5\right)^2\)

\(=12,4\cdot\dfrac{25}{4}-12,4\cdot\dfrac{25}{4}=0\)

b) \(32,125-\left(6,325+12,125\right)-\left(37+13,675\right)=32,125-6,325-12,125-37-13,675\)

\(=\left(32,125-12,125\right)-\left(6,325+13,675\right)-37=20-20-37=-37\)

\(4^x.8^x=1024\)

\(\Rightarrow32^x=1024\)

\(\Rightarrow32^2=1024\)

\(\Rightarrow x=2\)

2\(^x\).2\(^{x+3}\) = 144

2^2\(x\) = 144: 2³

4\(^x\) = 18

nếu \(x\) ≤ 2 ⇒ 4\(^x\) ≤ 4² = 16 < 18 (loại)

Nếu \(x\) ≥ 3 ⇒ 4\(^x\) ≥ 4³ = 64 > 18 (loại)

Vậy \(x\) \(\in\) \(\varnothing\)