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Lớp học online hay j mà lắm giáo viên thế🤔🤔 Lớp học của tri thức à?
a, \(\frac{2}{5}\)+ y = 1
y = 1 - \(\frac{2}{5}\)
y = \(\frac{3}{5}\)
b, \(\frac{6}{8}\)- y = \(\frac{3}{5}\)
y = \(\frac{6}{8}\)- \(\frac{3}{5}\)
y = \(\frac{3}{20}\)
c, y : 6 = \(\frac{7}{4}\)
y = \(\frac{7}{4}\)x 6
y = \(\frac{21}{2}\)
d, y - \(\frac{1}{3}\)= 5 x \(\frac{2}{3}\)
y - \(\frac{1}{3}\)= \(\frac{10}{3}\)
y = \(\frac{10}{3}\)+ \(\frac{1}{3}\)
y = \(\frac{11}{3}\)
a,2/5 + y = 1
y = 1 - 2/5
y = 3/5
b, 6/8 - y = 3/5
y = 6/8 - 3/5
y = 3/10
c, y / 6 = 7/4
y = 7/4 * 6
y = 15/2
d, y - 1/3 = 5 * 2/3
y - 1/3 = 10/3
y = 10/3 + 1/3
y = 11/3
\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)
y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)
y = \(\dfrac{245}{64}\)
2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)
\(\dfrac{12}{5}\): y = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)
y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)
y = \(\dfrac{15}{13}\)
\(\frac{7}{y}=\frac{1}{12}\)
\(=>\frac{7}{y}=\frac{7}{84}\)
\(=>y=84\)
\(\frac{9}{10}-y.\frac{3}{4}=\frac{2}{5}\)
\(y.\frac{3}{4}=\frac{9}{10}-\frac{2}{5}=\frac{1}{2}\)
\(y=\frac{1}{2}:\frac{3}{4}\)
\(y=\frac{1}{2}.\frac{4}{3}\)
\(y=\frac{2}{3}\)
\(\frac{5}{6}:\left[y+\frac{7}{9}\right]=\frac{3}{4}\)
\(y+\frac{7}{9}=\frac{5}{6}:\frac{3}{4}\)
\(y+\frac{7}{9}=\frac{5}{6}.\frac{4}{3}\)
\(y+\frac{7}{9}=\frac{10}{9}\)
\(y=\frac{10}{9}-\frac{7}{9}\)
\(y=\frac{3}{9}\)
\(y=\frac{1}{3}\)
1, 7/y=1/12
=>7.12=y
=>y=84
2, 9/10-y.3/4=2/5
=>y.3/4=9/10-2/5
=>y.3/4=1/2
=>y=1/2:3/4=2/3
3, 5/6[y+7/9]=3/4
=>y+7/9=5/6:3/4
=>y+7/9=10/9
=>y=10/9-7/9
=>y=1/3
\(y.3\dfrac{7}{12}=6\dfrac{1}{4}\)
\(y.\dfrac{43}{12}=\dfrac{25}{4}\)
\(y=\dfrac{25}{4}:\dfrac{43}{12}\)
\(y=\dfrac{25.12}{4.43}\)
\(y=\dfrac{75}{43}\)
còn a ơi