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11 tháng 4 2022

2/10 = 1/5

=> x = 1

3/5 x  = 1/4 + 1/3

3/5x = 7/12

x = 7/12 : 3/5

x = 35/36

1/5: x = 2/35

x = 1/5 : 2/35

x = 7/2

11 tháng 4 2022

\(a.x=\dfrac{5.2}{10}=1\\ b.x=\dfrac{3}{5}x=\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{3}{5}=\dfrac{7}{12}.\dfrac{5}{3}=\dfrac{35}{36}\\ c.\dfrac{1}{5}:x=\dfrac{2}{35}\\ x=\dfrac{1}{5}:\dfrac{2}{35}=\dfrac{1}{5}.\dfrac{35}{2}=\dfrac{7}{2}\)

26 tháng 4 2022

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

26 tháng 4 2022

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

1 tháng 4 2021

\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)

1 tháng 4 2021

\(\dfrac{1}{2}.x+\dfrac{1}{2}=\dfrac{5}{2}\Leftrightarrow\dfrac{1}{2}.x=\dfrac{5}{2}-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}.x=2\Leftrightarrow x=4\)

15 tháng 4 2022

a) \(\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)-3\dfrac{1}{2}=50\%\)

\(\dfrac{4}{5}x-\dfrac{4}{15}-\dfrac{7}{2}=\dfrac{1}{2}\)

\(\dfrac{4}{5}x=\dfrac{1}{2}+\dfrac{7}{2}+\dfrac{4}{15}\)

\(\dfrac{4}{5}x=\dfrac{64}{15}\)

\(x=\dfrac{64}{15}:\dfrac{4}{5}\\ x=\dfrac{16}{3}\)

b) \(8\dfrac{3}{5}.x-2\dfrac{1}{5}.x=16\%\)

\(\dfrac{43}{5}x-\dfrac{11}{5}x=\dfrac{4}{25}\)

\(\dfrac{32}{5}x=\dfrac{4}{25}\)

\(x=\dfrac{4}{25}:\dfrac{32}{5}\)

\(x=\dfrac{1}{40}\)

a: =>4x-5=0 hoặc 5/4x-2=0

=>x=5/4 hoặc x=2:5/4=2*4/5=8/5

b: =>(1/12+19/6-30,75)*x-8=102

=>-55/2x=110

=>x=-4

17 tháng 8 2021

b) \(3^x\cdot3^2+3^x=7290\)

\(3^x\left(3^2+1\right)=720\)

\(3^x\cdot10=7290\)

\(=>3^x=729=3^6\)

=> \(x=6\)

17 tháng 8 2021

ôi bạn ơi

mik ko bt

5 tháng 5 2022

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)

\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)

\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)

d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)

hay \(x=\dfrac{2}{3}\)