Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : 0,36⋅950+0,18⋅726⋅2+3⋅324⋅0,121+3+5+7+9+...+27+29+31−1521+3+5+7+9+...+27+29+31−1520,36⋅950+0,18⋅726⋅2+3⋅324⋅0,12
Đặt A = 0,36 . 950 + 0,18 . 726 . 2 + 3. 324 . 0,12
= 0,36 . 950 + 0,36 . 726 + 0,36 . 324
= 0,36(950 + 726 + 324) = 0,36.2000 = 720
Đặt B = (1 + 3 + 5 + 7 + 9 + ... + 27 + 29 + 31) - 152
Số số hạng : (31 - 1) : 2 + 1 = 16(số)
=> Tổng : (1 + 31).16 : 2 = 256
=> B = 256 - 152 = 104
Vậy ��=720104=9013BA=104720=1390
\(\left(\frac{5}{7}-y\right)x\frac{14}{5}=\frac{7}{10}+\frac{1}{2}\)
\(\left(\frac{5}{7}-y\right)x\frac{14}{5}=\frac{6}{5}\)
\(\frac{5}{7}-y=\frac{6}{5}:\frac{14}{5}\)
\(\frac{5}{7}-y=\frac{3}{7}\)
Y = \(\frac{5}{7}-\frac{3}{7}\)
Y = \(\frac{2}{7}\)
\(\left(\frac{5}{7}-y\right).\frac{14}{5}=\frac{7}{10}+\frac{1}{2}\)
\(\left(\frac{5}{7}-y\right).\frac{14}{5}=\frac{6}{5}\)
\(\frac{5}{7}-y=\frac{6}{5}\div\frac{14}{5}\)
\(\frac{5}{7}-y=\frac{3}{7}\)
\(y=\frac{5}{7}-\frac{3}{7}\)
\(y=\frac{2}{7}\)
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
x=\(\frac{4}{3}\)-\(\frac{3}{8}\)=\(\frac{23}{24}\)
x = \(\frac{4}{5}\)+ \(\frac{7}{6}\)=\(\frac{59}{30}\)
x = \(\frac{21}{8}\): \(\frac{1}{9}\) = \(\frac{189}{8}\)
x = \(\frac{8}{3}\)* 5 = \(\frac{40}{3}\)
chúc bạn học tốt ☻
\(\frac{1999\cdot2001-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)
\(=\frac{1999\cdot\left(2000+1\right)-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)
\(=\frac{1999\cdot2000+1999-1}{1998+1999.2000}\cdot\frac{7}{5}\)
\(=\frac{1999\cdot2000+1998}{1998+1999.2000}\cdot\frac{7}{5}=1\cdot\frac{7}{5}=\frac{7}{5}\)