Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, 2015 - 2 × X = 101
2 x X = 2015 - 101
2 x X = 1914
X = 1914 : 2
X = 957
1a. 2015 - 2 x X = 101
2 x X = 2015 - 101
X = 1914 : 2
X = 957
b. 7 - (11 + X - 13) : \(2\frac{2}{3}\)= 2
(11 + X - 13) : \(2\frac{2}{3}\) = 7 - 2
11 + X - 13 = 5 x \(2\frac{2}{3}\)
11 + X = \(\frac{40}{3}\)+ 13
X = \(\frac{79}{3}\)- 11
X = \(\frac{46}{3}\)
c. 2012 - 5 x X = 17
5 x X = 2012 - 17
X = 1995 : 5
X = 399
d. \(5\frac{3}{4}\)+ (X - 2) x \(\frac{1}{3}\)= \(17\frac{3}{4}\)
(X - 2) x \(\frac{1}{3}\) = \(17\frac{3}{4}\)- \(5\frac{3}{4}\)
X - 2 = 12 : \(\frac{1}{3}\)
X = 36 + 2
X = 38
e. 12 x (X - 5) + 73 = 20
12 x (X - 5) = 20 - 73
X - 5 = -53 : 12
X = \(\frac{-53}{12}\)+ 5
X = \(\frac{7}{12}\)
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
\(A=1+\frac{1}{2}+...+\frac{1}{16}\)
= \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{12}\right)+\left(\frac{1}{13}+...+\frac{1}{16}\right)\)
> \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
=\(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
=\(1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
= \(1+2\times\frac{13}{12}\)
= \(1+\frac{13}{6}\)
= \(1+2+\frac{1}{6}\)
= \(3+\frac{1}{6}\)>\(3\)
=> \(A>3+\frac{1}{6}>3\)
=> \(A>3+\frac{1}{6}>B\)
=> \(A>B\)