Bài 2:
1: 3x-1,7=1,5*3
=>3x-1,7=4,5
=>3x=4,5+1,7=6,2
=>\(x=\dfrac{6.2}{3}=\dfrac{62}{30}=\dfrac{31}{15}\)
2: \(2x-15=35\cdot\left(-1\right)^{10}\)
=>2x-15=35
=>2x=35+15=50
=>x=50/2=25
3: \(\dfrac{x}{-12}=\dfrac{-20}{6}\)
=>\(\dfrac{x}{12}=\dfrac{20}{6}=\dfrac{10}{3}\)
=>\(x=\dfrac{10}{3}\cdot12=10\cdot4=40\)
4: \(\dfrac{1}{4}x-50\%x=-1\dfrac{3}{4}\)
=>\(\dfrac{1}{4}x-\dfrac{1}{2}x=-\dfrac{7}{4}\)
=>\(-\dfrac{1}{4}x=-\dfrac{7}{4}\)
=>x=7
5: \(\dfrac{x}{1\cdot2}+\dfrac{x}{2\cdot3}+...+\dfrac{x}{99\cdot100}=\left(-10\right)^2\)
=>\(x\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100\)
=>\(x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)
=>\(x\left(1-\dfrac{1}{100}\right)=100\)
=>\(x\cdot\dfrac{99}{100}=100\)
=>\(x=100:\dfrac{99}{100}=\dfrac{10000}{99}\)
Bài 1:
1: \(\dfrac{1}{4}+\left(\dfrac{1}{2022}+\dfrac{3}{4}\right)\)
\(=\dfrac{1}{4}+\dfrac{1}{2022}+\dfrac{3}{4}\)
\(=1+\dfrac{1}{2022}=\dfrac{2023}{2022}\)
2: \(\dfrac{3}{19}\cdot\dfrac{-5}{17}+\dfrac{3}{19}\cdot\dfrac{-12}{17}+2022\dfrac{3}{19}\)
\(=\dfrac{3}{19}\left(-\dfrac{5}{17}-\dfrac{12}{17}\right)+2022+\dfrac{3}{19}\)
\(=-\dfrac{3}{19}+\dfrac{3}{19}+2022=2022\)
3: \(2,86\cdot4+3,14\cdot4-6,01\cdot5+3^2\cdot\left(-1\right)^{20}\)
\(=4\left(2,86+3,14\right)-30,05+9\)
\(=4\cdot6-30,05+9\)
=2,95
4: \(2,5\left(4,1-3-2,5+2\cdot7,2\right)+4,2:4\)
\(=2,5\left(1,1-2,5+14,4\right)+1,05\)
\(=2,5\cdot13+1,05=32,5+1,05=33,55\)
