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a , 5.6 + 5.7 / 5.8 + 20 = 5.6 + 5.7 / 5.8 + 5 . 4 = 5 . ( 6+7 ) / 5 . ( 8 + 4 ) = 6 + 7 / 8 + 4 = 13 / 12 8 . 9 + 4 .15 / 12 . 7 - 180 = 4 . 2 . 3 . 3 + 2 . 2 . 3 . / 4 . 3 . 7 - 180 = 4 . 2 . 3 . 3 + 2.2.3.5 / 3 . 4 . 7 - 3 . 2 . 2 . 3. 5 = 1 . 2 . 1 . 1 + 1 . 1 . 1. 1 / 1 . 1 . 7 - 1 . 1 . 1 . 1 .1 = 3 / 6 = 1/2 b , 2^5 . 7 +2^5 / 2^5 . 5^2 - 2^5 .3 = 2^5 . ( 7 + 1) / 2^5 ( 5^2 - 3 ) = 7+1 / 5^2 - 3 = 8 / 22 = 4 / 11 3^4 . 5 - 3^6 / 3^4 . 13 + 3^4 = 3^4 . 5 - 3^4 . 3^2 / 3^4 . 13 + 3^4 = 3^4 . ( 5 - 3^2 ) / 3^4 . ( 13 + 1 ) = 5 - 3^2 / 13 + 1 = -4 / 14 = -2 / 12
Ta có: \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\) = \(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\) = \(\frac{2^5.8}{2^5.22}\) = \(\frac{8}{22}\) =\(\frac{56}{154}\)
\(\frac{3^4.5-3^6}{3^4.13+3^4}\) = \(\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}\) = \(\frac{3^4.\left(-4\right)}{3^4.14}\) = \(\frac{-4}{14}\)= \(\frac{-44}{154}\)
\(a,\left(-25\right).\left(75-45\right)-75.\left(45-25\right)\)
\(=\left(-25\right).20-75.20\)
\(=20.\left(-25-75\right)\)
\(=20.\left(-100\right)\)
\(=-2000\)
\(b,\frac{18.12-48.15}{-3.270-3.30}\)
\(=\frac{18.12-12.4.15}{-3.270-3.30}\)
\(=\frac{12.\left(18-60\right)}{-3.\left(270+30\right)}\)
\(=\frac{12.\left(-42\right)}{-3.300}\)
\(=\frac{14}{25}\)
\(c,\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(=\frac{2^5.\left(7+1\right)}{2^5\left(25-3\right)}\)
\(=\frac{2^5.8}{2^5.22}\)
\(=\frac{8}{22}\)
\(=\frac{4}{11}\)
Câu 1:
\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4
Vậy A = 4
Câu 2:
\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)
\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)
\(B=1000-3.\left\{400-273\right\}\)
\(B=1000-3.127\)
\(B=1000-381\)
\(B=619\)
Vậy B = 619
\(a,\frac{5\cdot6+5\cdot7}{5\cdot8+20}=\frac{\left(5\cdot6+5\cdot7\right)\div2}{\left(5.8+20\right)\div2}=\frac{13}{12}\)
\(\frac{8.9-4.15}{12.7-180}=\frac{12.6-12.5}{12.7-12.15}=\frac{1}{-8}=\frac{-1}{8}\)
\(Ta\)\(co\)\(\frac{13}{12}=\frac{13.2}{12.2}=\frac{26}{24}\)
\(\frac{-1}{8}=\frac{3\left(-1\right)}{3.8}=\frac{-3}{24}\)
Phần b bạn tính ra rồi làm tương tự phần a nha chúc bạn học giỏi!!!
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
dấu / là phần nhé. bạn có thể xem bài có dấu phần ở : Câu hỏi của Nguyễn Thị Hoài Anh
A)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
=1-\(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=1-\(\frac{1}{101}\)
=\(\frac{100}{101}\)
B) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{1}{99.101}\)
=5.(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{2}{2}.\)(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))
=\(\frac{5}{2}.\frac{100}{101}=\frac{250}{100}\)
Chúc bạn học tốt
\(\frac{2^5.7+2^5}{2^5+5^2-2^5.3}\)
\(=\frac{2^5.8}{5^2.3}\)
\(=\frac{32.8}{25.3}\)
\(=\frac{256}{75}\)