B = ( 24 - 23 1/9). ( - 3/4 )2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. Tính:
a) 23 + 24 - ( 65 x 2 - 4 ) x 4 + 66 + 44 - ( 54 : 9 x 9 : 9 + 45 ) : 1 + 1 - 1 x ( 23 + 23 - 23 x 2 +1 ) = 51
b) 4444444444444444444444444444444444444444444444444444444444444444444 : 1
= 4444444444444444444444444444444444444444444444444444444444444444444
c) 30 = 0
d) 1/1 = 1
e)
a) - 530
b) 44444444444444444444444444444444444444444444444444444444444444444444
c) 0
d) 1
e) Em là em
\(a,\dfrac{-1}{7}+\dfrac{8}{13}+\dfrac{-6}{7}-1\dfrac{23}{24}-\dfrac{-5}{13}\\ =\dfrac{-1}{7}+\dfrac{8}{13}+\dfrac{-6}{7}-\dfrac{47}{24}+\dfrac{5}{13}\\ =\left(\dfrac{-1}{7}+\dfrac{-6}{7}\right)+\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{47}{24}\\ =-1+1-\dfrac{47}{24}\\ =0-\dfrac{47}{24}\\ =\dfrac{-47}{24}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}\\ =7\dfrac{4}{9}+4\dfrac{7}{11}-3\dfrac{4}{9}\\ =\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{7}{11}\\ =4+4\dfrac{7}{11}\\ =\dfrac{44}{11}+\dfrac{51}{11}\\ =\dfrac{95}{11}\)
Giải:
a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\)
\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\)
\(=\dfrac{136}{9}-\dfrac{11}{2}\)
\(=\dfrac{173}{18}\)
b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\)
\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\)
\(=\dfrac{13}{9}.\dfrac{3}{4}\)
\(=\dfrac{13}{12}\)
c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\)
\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=1+1\)
\(=2\)
d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\)
\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\)
\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\)
\(=\dfrac{53}{10}\)
e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-2}{5}\)
\(=1\)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)