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\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\)\(\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(15+5x=50\)
\(5x=35\)
\(x=7\)
Vậy \(x=7\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\Rightarrow1+x+2+x+3+x+4+x+5+x=50\)
\(\Rightarrow\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(\Rightarrow15+5x=50\)
\(\Rightarrow5x=50-15\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=35:5\)
\(\Rightarrow x=7\).
\(320\div x-10=5\cdot48\div24\)
\(320\div x-10=240\div24\)
\(320\div x-10=10\)
\(320\div x=10+10\)
\(320\div x=20\)
\(x=320\div20\)
\(x=16\)
a) \(\frac{1}{2}\times x-3=6\)
=> \(\frac{1}{2}\times x=6+3\)
=> \(\frac{1}{2}\times x=9\)
=>\(x=9:\frac{1}{2}\)
=> \(x=18\)
b) \(2:x=\frac{2}{5}--\frac{1}{10}\)
=> \(2:x=\frac{2}{5}+\frac{1}{10}\)
=> \(2:x=\frac{1}{2}\)
=> \(x=2:\frac{1}{2}\)
=> \(x=4\)
c) \(25-\left(2\frac{1}{2}+x\right)=10\)
=> \(2\frac{1}{2}+x=25-10\)
=> \(\frac{5}{2}+x=15\)
=>\(x=15-\frac{5}{2}\)
=> \(x=\frac{25}{2}\)
d) \(\left(x-\frac{3}{4}\right)\times3-45:9=10\)
=> \(\left(x-\frac{3}{4}\right)\times3-5=10\)
=> \(\left(x-\frac{3}{4}\right)\times3=10+5\)
=> \(\left(x-\frac{3}{4}\right)\times3=15\)
=> \(\left(x-\frac{3}{4}\right)=15:3\)
=> \(\left(x-\frac{3}{4}\right)=5\)
=> \(x=5+\frac{3}{4}\)
=> \(x=\frac{23}{4}\)
\(a,\frac{1}{2}.x-3=6\Rightarrow\frac{x}{2}=9\Rightarrow x=18\)
\(b,2:x=\frac{2}{5}-\frac{1}{10}\Rightarrow\frac{2}{x}=\frac{9}{10}\Rightarrow x=\frac{2.10}{9}=\frac{20}{9}\)
\(c,25-\left(2\frac{1}{2}+x\right)=10\Rightarrow25-\frac{5}{2}+x=10\Rightarrow x=10+\frac{5}{2}-25=-\frac{25}{2}\)
\(d,\left(x-\frac{3}{4}\right).3-45:9=10\Rightarrow\left(x-\frac{3}{4}\right).3-5=10\Rightarrow\left(x-\frac{3}{4}\right).3=15\Rightarrow x-\frac{3}{4}=5\Rightarrow x=\frac{23}{4}\)
\(x(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\) \(5\frac{1}{2}\)
\(x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\cdot\frac{11}{12}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{11}{12}\)
\(x=6\)
Vậy x = 6
\(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+...+\frac{x}{132}=5\frac{1}{2}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}=\frac{11}{2}\right)\)
\(\Rightarrow x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(\Rightarrow x.\frac{11}{12}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{11}{12}=1\)
Vậy x = 1
câu hỏi tương tự
x−25,34=45:5
x-25,34=9
x=9 + 25,34
x=34,34
x+54,36=99,87−13,6
x+54,36=86,27
x=86,27- 54,36
x=29,91