Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)
= 22 \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{110}{3}\)
\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)
= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5
= \(\dfrac{28}{15}\) + 5
= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)
= \(\dfrac{103}{15}\)
\(A=\dfrac{5}{2}\times\dfrac{2}{3}+\dfrac{1}{3}\)
\(=\dfrac{5\times2}{2\times3}+\dfrac{1}{3}\)
\(=\dfrac{10}{6}+\dfrac{1}{3}\)
\(=\dfrac{5}{3}+\dfrac{1}{3}\)
\(=\dfrac{6}{3}\)
\(=2\)
a: \(=\dfrac{3}{5}\cdot\dfrac{5}{4}+\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{9+4}{12}=\dfrac{13}{12}\)
b: \(\Leftrightarrow x\cdot\dfrac{5}{4}=\dfrac{5}{8}\)
hay x=1/2
a: \(x:5=\dfrac{15}{14}\\ \Rightarrow x=\dfrac{15}{70}=\dfrac{3}{14}\)
(1+2+3+4+...+96+97+98+99):5
Đặt 1+2+3+4+...+96+97+98+99=A1+2+3+4+...+96+97+98+99=S
Số số hạng của S là:
(99−1):1+1=99(99-1):1+1=99
Tổng của S là:
(99+1).99:2=4950(99+1).99:2=4950
→(1+2+3+4+...+96+97+98+99):5→(1+2+3+4+...+96+97+98+99):5
=4950:5=990
S=(1+2+3+4+...+96+97+98+99):5
S=(99x(99+1):2):5
S=(99x100:2):5
S=(9900:2):5
S=4950:5
S=990
a: Thay a=9 và b=15 vào P, ta được:
\(P=\left(9+1\right)\cdot2+\left(15+1\right)\cdot3\)
\(=10\cdot2+16\cdot3=20+48=68\)
b: \(m=2\cdot a+3\cdot b+5=2\cdot9+3\cdot15+5=68\)
mà P=68
nên P=m
1/10 + 3/5 : 2/3
=1/10 + 9/10
=10/10
=1
\((\)6/8 -2/6\()\) x 1/2
=18/48 X 1/2
=3/7 x 1/2
=2/14
\(\dfrac{1}{10}\) + \(\dfrac{3}{5}\):\(\dfrac{2}{3}\)
= \(\dfrac{1}{10}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{3}{2}\)
= \(\dfrac{1}{10}\) + \(\dfrac{9}{10}\)
= \(\dfrac{10}{10}\)
= 1
( \(\dfrac{6}{8}\) - \(\dfrac{2}{6}\)) \(\times\) \(\dfrac{1}{2}\)
=( \(\dfrac{18}{24}\) - \(\dfrac{8}{24}\)) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{10}{24}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{5}{24}\)
Đáp án: 1