Tính:
3/4 : 3
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Chuyển các phép tính sau thành phép nhân rồi tính:
a) 3 + 3 +3 + 3 + 3
= 3 x 5
= 15
b) 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4
= 3 x 4 + 4 x 4
= 12 + 16
= 28
Chúc bạn học tốt !
5,5 + \(\dfrac{3}{4}\) - 5 + \(\dfrac{1}{4}\)
(5,5 - 5) + (\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\))
= 0,5 + 1
= 1,5
\(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) : \(\dfrac{3}{2}\)
= \(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) x \(\dfrac{2}{3}\)
= (\(\dfrac{5}{2}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)
= (\(\dfrac{10}{4}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)
= \(\dfrac{11}{4}\) x \(\dfrac{2}{3}\)
= \(\dfrac{11}{6}\)
Tính \(x\):
435 - [\(x\) + 16] = 425 : 17
435 - [\(x\) + 16] = 25
[\(x\) + 16] = 435 - 25
\(x\) + 16 = 410
\(x\) = 410 - 16
\(x\) = 394
Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
Bài 1:
a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)
= \(\dfrac{29}{10}\)
b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)
= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)
= \(\dfrac{19}{20}\)
c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
Bài 2:
3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\)
= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)
= \(\dfrac{28}{5}\)
b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)
Lời giải chi tiết:
3 + 1 = 4 | 4 – 2 = 2 | 1 + 2 = 3 |
4 – 3 = 1 | 3 – 2 = 1 | 3 – 1 = 2 |
4 – 1 = 3 | 4 – 3 = 1 | 3 – 2 = 1 |
\(\dfrac{3}{4}:3\)
\(=\dfrac{3}{4}\times\dfrac{1}{3}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
Chúc bạn học tốt
\(\dfrac{3}{4}:3\)
\(=\dfrac{3}{4}\times\dfrac{1}{3}\)
\(=\dfrac{1}{4}\)