Tìm x: x : 7 11 = 22
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a, \(\Leftrightarrow x=1-\dfrac{2}{9}=\dfrac{7}{9}\)
b, \(\Leftrightarrow x=\dfrac{22.7}{11}=2.7=14\)
x-7-(11-3x)=22
x-7-11+3x=22
x+3x=22+7+11
4x=40
x=40:4=10
Vậy x cần tìm là 10
\(\dfrac{\dfrac{22}{15}}{11-x}\) + \(\dfrac{2}{3}\) = \(\dfrac{7}{5}\)
\(\dfrac{\dfrac{22}{15}}{11-x}\) = \(\dfrac{7}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{\dfrac{22}{15}}{11-x}\) = \(\dfrac{11}{15}\)
11 - \(x\) = \(\dfrac{22}{15}\) : \(\dfrac{11}{15}\)
11 - \(x\) = 2
\(x\) = 11 - 2
\(x\) = 9
Giải luôn nhé:
7x - x : 1/5 = 28
=> 7x - x.5 = 28
=> 2x = 28
=> x = 14
c) 16 - x + 14 = 22
16 - x = 36
x = -20
\(\frac{22}{7}:\left(11-x\right)+\frac{2}{3}=\frac{7}{5}\)
\(\frac{22}{7}:\left(11-x\right)=\frac{11}{15}\)
\(11-x=\frac{30}{7}\)
\(x=\frac{47}{7}\)
Vậy \(x=\frac{47}{7}\)
\(\frac{22}{7}\div\left(11-x\right)+\frac{2}{3}=\frac{7}{5}\)
\(\Leftrightarrow\frac{2}{7}-\frac{22}{7x}+\frac{2}{3}=\frac{7}{5}\)
\(\Rightarrow\frac{30x}{105x}+\frac{330}{105x}+\frac{70x}{105x}=\frac{147}{105x}\)
\(\Rightarrow100x+330=147\)
\(\Rightarrow100x=-153\)
\(\Rightarrow x=-1,53\)