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a:
Sửa đề: 19/17-19/49+19/131
\(=\dfrac{19\left(\dfrac{1}{17}-\dfrac{1}{49}-\dfrac{1}{131}\right)}{3\left(\dfrac{1}{17}-\dfrac{1}{49}-\dfrac{1}{131}\right)}=\dfrac{19}{3}\)
b: \(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}}{\left(1+\dfrac{2019}{2}\right)+\left(1+\dfrac{2018}{3}\right)+...+\left(1+\dfrac{1}{2020}\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)
3²ˣ⁺¹ - 20 = 7
3²ˣ⁺¹ = 7 + 20
3²ˣ⁺¹ = 27
3²ˣ⁺¹ = 3³
2x + 1 = 3
2x = 3 - 1
2x = 2
x = 2 : 2
x = 1
\(-\left(-x\right):\left(-15\right).2=16\)
\(\Rightarrow x:\left(-15\right)=8\)
\(\Rightarrow x=-120\)
Vậy: \(x=-120\)
\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)
\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)
\(\Leftrightarrow2x+8=-x+11\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
x. ( 4 + 40 ) = 101 . 11
x. 44 = 1111
x = 1111 : 44
x = 25,25
x04 + 40x = 101.11
=> 100x + 4 + 400 + x = 1111
=> 101x + 404 = 1111
=> x = 7
\(\Leftrightarrow2x+1\in\left\{1;7\right\}\)
hay \(x\in\left\{0;3\right\}\)
Vì \(x\in N\Rightarrow2x+1\in N\) và 2x+1 lẻ
\(14⋮2x+1\Rightarrow2x+1\inƯ\left(14\right)=\left\{1;7\right\}\)
Ta có bảng:
Vậy \(x\in\left\{0;3\right\}\)