Các câu hỏi tương tự
\(\left(\frac{1}{25.26}+\frac{1}{26.27}+............+\frac{1}{29.30}\right).150+103\div\left[1.03+\left(x+1\right)\right]=22\)
(1/25 x 26 + 1/26x27+...+1/29x30)x150 + 1,03: [1,03x(X-1)] =22
18 tháng 4 2022 lúc 20:45
2,07 + (-7,36)-(-8,97)+1,03-7,64
1,01+1,02+1,03+...+9,98+9,99+10
\(_{A=\left(\frac{1}{26.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x+x:\frac{1}{3}=4}\)
Ai giúp em vs!
Cho A = 1,01 + 1,02 + 1,03 + ... + 9,98 + 9,99 + 10
và B = \(2-\frac{5}{3}+\frac{7}{6}-\frac{9}{10}+\frac{11}{15}-\frac{13}{21}+\frac{15}{28}-\frac{17}{36}+\frac{19}{45}\)
Tính 2A + \(\frac{455}{3}\)B
a) 60,7 + 25,5 - 38,7
b) (-9,207) + 3,8 + (-1,5030) - 2,8
c) (-12,5) + 17,55 + (-3,5) - (-2,45)
d) 2,07 + (-7,36) - (-8,97) + 1,03 - 7,64
e) (2,07 + 3,005) - (12,005 - 4,23)
f) 4,35- (2,67 - 1,65) + (3,54 - 6,33
Tìm x
a/\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}+x\div\frac{1}{3}=-4\)
b/\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right)\left(3x+5\right)}=\frac{3}{20}\)
Tìm x
a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)