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\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)
\(\dfrac{1}{6}+x=\dfrac{14}{5}\)
\(=>x=\dfrac{79}{30}\)
\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)
\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)
\(\dfrac{32}{35}:x=\dfrac{11}{6}\)
\(x=\dfrac{192}{385}\)
\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)
\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)
\(x:\dfrac{3}{4}=\dfrac{38}{30}\)
\(=>x=\dfrac{19}{20}\)
\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)
( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ( x + 7 ) + ( x + 9 ) + ( x + 11 ) + ( x + 13 ) = 119
( x + x + x + x + x + x + x ) + ( 1 + 3 + 5 + 7 + 9 + 11 + 13 ) = 119
7x + 49 = 119
7x = 119 - 49
7x = 70
x = 70 : 7
x = 10
Câu đầu em xem lại đề bài sao có hai dấu bằng.
Câu 2:
\(\dfrac{3}{2}\) \(\times\)y - \(\dfrac{3}{4}\) \(\times\)y + y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{3}{2}\) - \(\dfrac{3}{4}\) + 1) = \(\dfrac{4}{5}\)
y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)
y \(\times\) \(\dfrac{7}{4}\) = \(\dfrac{4}{5}\)
y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)
y = \(\dfrac{16}{35}\)
11/3+11/5.y-4/5=24/5
88/15.y=24/5-4/5
88/15.y=20/5
88/15.y=4
y=4:88/15
y=15/22
11/3 + 11/5 x y - 4/5 = 24/5
11/3 + 11/5 x y = 24/5 + 4/5
11/3 + 11/5 x y = 28/5
11/5 x y = 28/5 - 11/3
11/5 x y = 29/15
y = 29/15 : 11/5
y = 29/33