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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) \(\left(x-\frac{3}{4}\right)\times3-45:9=10\)
=> \(\left(x-\frac{3}{4}\right)\times3-5=10\)
=> \(\left(x-\frac{3}{4}\right)\times3=10+5\)
=> \(\left(x-\frac{3}{4}\right)\times3=15\)
=> \(\left(x-\frac{3}{4}\right)=15:3\)
=> \(\left(x-\frac{3}{4}\right)=5\)
=> \(x=5+\frac{3}{4}\)
=> \(x=\frac{23}{4}\)
\(a,\frac{1}{2}.x-3=6\Rightarrow\frac{x}{2}=9\Rightarrow x=18\)
\(b,2:x=\frac{2}{5}-\frac{1}{10}\Rightarrow\frac{2}{x}=\frac{9}{10}\Rightarrow x=\frac{2.10}{9}=\frac{20}{9}\)
\(c,25-\left(2\frac{1}{2}+x\right)=10\Rightarrow25-\frac{5}{2}+x=10\Rightarrow x=10+\frac{5}{2}-25=-\frac{25}{2}\)
\(d,\left(x-\frac{3}{4}\right).3-45:9=10\Rightarrow\left(x-\frac{3}{4}\right).3-5=10\Rightarrow\left(x-\frac{3}{4}\right).3=15\Rightarrow x-\frac{3}{4}=5\Rightarrow x=\frac{23}{4}\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
Ta có : \(\frac{1}{4}+\frac{1}{3}:\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow\frac{1}{3}:\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}:\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}:\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\times\frac{3}{2}\)
\(\frac{1}{x}=\frac{1}{2}\)
=> x = 2
a) \(\frac{x\div3-16}{2}+21=38\)
\(\frac{x\div3-16}{2}=38+21\)
\(\frac{x\div3-16}{2}=59\)
\(x\div3-16=59.2\)
\(x\div3-16=118\)
\(x\div3=118+16\)
\(x\div3=134\)
\(x=134.3\)
\(x=402\)
b) \(\frac{1}{4}+\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\div\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{2}\)
Vậy x = ....
1)
a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)
\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)
\(x+\frac{127}{128}=5\)
\(x=5-\frac{127}{128}=\frac{513}{128}\)
b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)
\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)
\(x+\frac{2186}{2187}=3\)
\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)
2)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)
\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)
\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)
\(=8+2=10\)
c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)
\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)
\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)
\(=11+\frac{257}{120}=\frac{1577}{120}\)
3) Gọi số đó là x. Theo đề ta có :
\(\frac{16-x}{21+x}=\frac{5}{7}\)
\(7\left(16-x\right)=5\left(21+x\right)\)
\(112-7x=105+5x\)
\(112-105=7x-5x\)
\(7=2x\)
\(x=\frac{7}{2}=3,5\) ( vô lí )
Vậy không có số tự nhiên để thõa mãn điều kiện trên.
dạng chuỗi nha bạn
ko hiểu thì tích cho mình là mình giải cho