Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3-x\times\frac{1}{2}=\frac{3}{4}\)
\(x\times\frac{1}{2}=3-\frac{3}{4}\)
\(x\times\frac{1}{2}=\frac{12}{4}-\frac{3}{4}\)
\(x\times\frac{1}{2}=\frac{9}{4}\)
\(x=\frac{9}{4}\div\frac{1}{2}\)
\(x=\frac{9}{4}\times\frac{2}{1}\)
\(x=\frac{18}{4}=\frac{9}{2}\)
Vậy x = 9/2
ik r mk làm tiếp cho
\(\left(a\right)37064-64\left(82+\frac{42966}{217}\right)\)
\(=37064-64\left(82+198\right)\)
\(=37064-64\cdot280\)
\(=30764-17920=19144\)
\(\left(b\right)320-\left(120,5+95,25+5,25\right)+\frac{84}{12}\cdot12,5\)
\(=320-\left(120,5+100,5\right)+7\cdot12,5\)
\(=320-221+87,5=11,5\)
\(\left(c\right)\left(4\frac{2}{5}+2\frac{3}{7}\right)-\left(2\frac{2}{5}-5\frac{4}{7}\right)\)
\(=4\frac{2}{5}-2\frac{2}{5}+2\frac{3}{7}+5\frac{4}{7}\)
\(=2+8=10\)
\(\left(d\right)\left(\frac{1998}{18}-\frac{1443}{13}\right)\left(16996-\frac{1110}{30}\cdot305\right)\)
\(=\left(111-111\right)\left(16996-\frac{1110}{30}\cdot305\right)\)
\(=0\left(16996-\frac{1110}{30}\cdot305\right)=0\)
\(\left(e\right)5\frac{3}{5}+1,75+6\frac{1}{8}+4\frac{1}{4}+3,875+3,4\)
\(=5\frac{3}{5}+1\frac{3}{4}+6\frac{1}{8}+4\frac{1}{4}+3\frac{7}{8}+3\frac{2}{5}\)
\(=5\frac{3}{5}+3\frac{2}{5}+1\frac{3}{4}+4\frac{1}{4}+6\frac{1}{8}+3\frac{7}{8}\)
\(=9+6+10=25\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
a)12:x+4=3+25:5
12:x+4=3+5
12:x+4=8
12:x=8-4
12:x=4
x=12:4
x=3
vậy x = 3
a)12 :x+4=3+25:5
12:x+4=8
12:x=8-4
12:x=2
x=12:2
x=6
bn tự giải câu B nhé
câu C ko rõ đầu bài mik ko giaỉ được
d) (x-1/4).5/3=7/4-1/2
(x-1/4). 5/3=5/4
(x-1/4)=5/4:5/3
(x-1/4)=3/4
x =3/4-1/4
x =2/4
bn cố gắng giải nốt 2 câu còn lại hoặc bn ghi rõ đề bài con C mik giải lại cho
a) 12 : x + 5 = 33
12 : x = 33 - 5
12 : x = 28
x = 12 : 28
x = 1/4
Vậy x = 1/4
b) 5*(x + 19) = 170 - 50
5*(x+19) = 120
x + 19 = 120 : 5
x + 19 = 24
x = 24 - 19
x = 5
Vậy x = 5
c) 11*(x - 6) - 11 = 4*x
11*(x-7) = 4x
11x - 77 = 4x
11x - 4x = 77
7x = 77
x = 11
Vậy x = 11
b, 125 - 25:3 x 12 = 125 - \(\frac{25}{3}\)x 12 = 125 - 100 = 25
a, 50% + \(\frac{7}{12}\)- \(\frac{1}{2}\) = \(\frac{50}{100}+\frac{7}{12}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}+\frac{7}{12}=0+\frac{7}{12}=\frac{7}{12}\)