Question

[sophia]

\(\left(\frac{1}{4}+35\%+0,65+75\%\right)X\left(1\frac{6}{3}+4\frac{8}{24}\right)\)

X là nhân

Answer 1

[Hatsune Miku]

\(\left(\frac{1}{4}+35\%+0,65+75\%\right)x\)\(\left(1\frac{6}{3}+4\frac{8}{24}\right)\)

\(=\left(\frac{1}{4}+\frac{7}{20}+\frac{13}{20}+\frac{3}{4}\right)\)\(x\left(3+\frac{13}{3}\right)\)

\(=\left(\frac{5}{20}+\frac{7}{20}+\frac{13}{20}+\frac{15}{20}\right)\)\(x7\)

\(=2x7\)

\(=14\)