\(\dfrac{4}{7}< \dfrac{x}{10}< \dfrac{5}{7}\)

\(\Rightarrow\dfrac{4.10}{7.10}< \dfrac{7.x}{10.7}< \dfrac{5.10}{7.10}\)

\(\Rightarrow\dfrac{40}{70}< \dfrac{7.x}{70}< \dfrac{50}{70}\)

\(\Rightarrow7.x\in\left\{41;42;43;...49\right\}\)

\(\Rightarrow7.x\in\left\{42;49\right\}\) \(\left(x\in N\Rightarrow7.x⋮7\right)\)

\(7.x=42\Leftrightarrow x=6\)

\(7.x=49\Leftrightarrow x=7\)

\(35-\left[\left(2x-3\right)^2:7\right]=28\)

\(\Rightarrow\left[\left(2x-3\right)^2:7\right]=35-28\)

\(\Rightarrow\left(2x-3\right)^2:7=7\)

\(\Rightarrow\left(2x-3\right)^2=1\)

\(\Rightarrow2x-3=\pm1\)

\(\Rightarrow x=2\) hay \(x=1\)

Số người bổ sung 15+5=20 (người).

Số ngày ngôi nhà làm xong khi có 20 người là :

20x15:20= 15 (ngày)

\(\dfrac{11}{31}=\dfrac{11x311}{31x311}=\dfrac{3421}{31x311}\)

\(\dfrac{111}{311}=\dfrac{111x31}{31x311}=\dfrac{3441}{31x311}\)

mà \(\dfrac{3421}{31x311}< \dfrac{3441}{31x311}\)

\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}\)

a) \(x+10=20\Leftrightarrow x=10\)

b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)

c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)

d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)

e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)

f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)

g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)

h) \(2.\left(x+9+10+11\right)=4.12.25\)

\(\Leftrightarrow x+30=2.12.25\)

\(\Leftrightarrow x=600-30\)

\(\Leftrightarrow x=570\)

\(MSC=254.1232\)

\(\dfrac{253}{-254}=\dfrac{-253}{254}=\dfrac{-253.1232}{254.1232}=\dfrac{\text{-311696}}{254.1232}\)

\(\dfrac{-1234}{1232}=\dfrac{-1234.254}{254.1232}=\dfrac{\text{-313436}}{254.1232}\)

mà \(\dfrac{\text{-313436}}{254.1232}< \dfrac{\text{-311696}}{254.1232}\)

\(\Rightarrow\dfrac{-1234}{1232}< \dfrac{253}{-254}\)

\(\)

\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)

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

\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)

\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)

\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)

\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)

mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)

\(\Rightarrow A< B\)

333...333 (100 số)x333.333 (100 số)

= 3.111...111(100 số 1)x3.111...111 (100 số 1)

= 9.(111...111)² (100 số)

Đặt A₁₀₀=9.(111...111)² (100 số)

Ta thấy :

A₂=11²=121 ⇒121.9=1089

A₃=111²=12321 ⇒12321.9=110889

A₁₀=11..11² (10 chữ số)=1234567900987654321 ⇒A₁₀.9=11111111108888888889

A₁₅=11..11² (15 chữ số)=12345679012345654320987654321 ⇒A₁₀.9=11..1088..89 (14 số 1, 1 sô 0, 14 số 8, 1 số 9)

...............

⇒ A₁₀₀.9=11..1088..89 (99 số 1, 1 sô 0, 99 số 8, 1 số 9)

\(\dfrac{1}{1.4}+\dfrac{1}{4.2}+\dfrac{1}{2.8}+\dfrac{1}{8.4}+\dfrac{1}{4.16}+\dfrac{1}{16.8}\)

= \(\dfrac{1}{4}.\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)

= \(\dfrac{1}{4}.\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}\right)\)

= \(\dfrac{1}{4}.\dfrac{3}{2}\left(1+\dfrac{1}{4}+\dfrac{1}{16}\right)\)

= \(\dfrac{1}{4}.\dfrac{3}{2}\left(\dfrac{16}{16}+\dfrac{4}{16}+\dfrac{1}{16}\right)\)

= \(\dfrac{3}{8}.\dfrac{21}{16}\)

= \(\dfrac{63}{128}\)

\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

= \(\dfrac{337.2023}{2}\)

= \(\dfrac{\text{681751}}{2}\)