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\(x.\frac{1}{2}=\frac{1}{2}+\frac{2}{4}\)
\(x.\frac{1}{2}=\frac{3}{4}\)
\(x=\frac{3}{4}:\frac{1}{2}\)
\(x=\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{2}{4}+\frac{1}{2}\)
\(x-\frac{1}{2}=1\)
\(x=1+\frac{1}{2}\)
\(x=\frac{3}{2}\)
\(x.\frac{1}{2}=2\) ; \(x+2=\frac{1}{2}+\frac{1}{2}\)
\(x=2:\frac{1}{2}\) \(x+2=\frac{2}{4}\)
\(x=4\) \(x=\frac{2}{4}-2\)
\(x=\frac{-6}{4}\)
\(x.\frac{1}{2}=2\)
\(x=2:\frac{1}{2}\)
\(x=4\)
\(x+2=\frac{1}{2}+\frac{1}{2}\)
\(x+2=1\)
\(x=1-2\)
\(khong\)\(co\)\(so\)\(tu\)\(nhien\)\(thoa\)\(man\)
\(k\) \(mk\)
Mình làm cách lớp 6 nha bạn, cách lớp 4 mình không biết.
\(\frac{1}{2}.x+\frac{1}{5}.x=\frac{2}{5}\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{2}{5}\)
\(\Rightarrow x.\frac{7}{10}=\frac{2}{5}\)
Giờ thành tìm x rồi!
\(x=\frac{7.5}{2.10}\)
\(x=1\frac{3}{4}\)
Xong!
\(\frac{1}{2}\times x+\frac{1}{5}\times x=\frac{2}{5}\)
\(x\times\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{2}{5}\)
\(x\times\frac{7}{10}=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{7}{10}\)
\(x=\frac{4}{7}\)
\(x-\frac{1}{2}=1+\frac{2}{1}\)
\(x-\frac{1}{2}=3\)
\(x=3+\frac{1}{2}\)
\(x=\frac{7}{2}\)
x- 1/2= 1+2/1
=>x-1/2=1+2
=>x-1/2=3
=>x=3+1/2
=>x=6/2+1/2
=>x=7/2
\(2\frac{1}{2}-x=1\frac{2}{3}\)
\(x=2\frac{1}{2}-1\frac{2}{3}\)
\(x=\frac{5}{6}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(4x+\frac{15}{16}=\frac{23}{16}\)
\(4x=\frac{1}{2}\)
\(x=\frac{1}{8}\)
Vậy \(x=\frac{1}{8}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Rightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Rightarrow5x+\frac{15}{32}=\frac{23}{16}\)
\(\Rightarrow5x=\frac{23}{16}-\frac{15}{32}\)
\(\Rightarrow5x=\frac{31}{32}\)
\(\Rightarrow x=\frac{31}{32}.\frac{1}{5}=\frac{31}{160}\)
\(x+3\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}-3\frac{1}{2}\)
\(x=-3\)
\(x=2\frac{1}{2}-1\frac{1}{2}=1\)
\(2\frac{1}{2}-x=1\frac{1}{2}\)
\(x=2\frac{1}{2}-1\frac{1}{2}\)
\(x=1\)