2 023 + 25^2 : 5^3 + 27
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TB
2
T
17 tháng 9 2023
\(a, 2023 - 25^2 : 5^3 + 27\)
\(=\left(2023+27\right)-\left(5^2\right)^2:5^3\)
\(=2050-5^4:5^3\)
\(=2050-5\)
\(=2045\)
___
\(b,60:\left[7\cdot\left(11^2-20\cdot6\right)+5\right]\)
\(=60:\left[7\cdot\left(121-120\right)+5\right]\)
\(=60:\left(7\cdot1+5\right)\)
\(=60:\left(7+5\right)\)
\(=60:12\)
\(=5\)
#Toru
VN
Võ Ngọc Phương
VIP
17 tháng 9 2023
a) \(2023-25^2:5^3+27\)
\(=2023-5^4:5^3+27\)
\(=2023-5+27\)
\(=2018+27\)
\(=2045\)
b) \(60:\left[7.\left(11^2-20.6\right)+5\right]\)
\(=60:\left[7.\left(121-120\right)+5\right]\)
\(=60:\left[7.1+5\right]\)
\(=60:\left[7+5\right]\)
\(=60:12\)
\(=5\)
\(#WendyDang\)
NV
1
13 tháng 8 2017
\(\dfrac{-2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{19.21}-\dfrac{2}{21.23}-\dfrac{2}{23.25}-\dfrac{2}{25.27}-\dfrac{1}{27}\)\(=\dfrac{-2}{1.3}+\dfrac{-2}{3.5}+\dfrac{-2}{5.7}+...+\dfrac{-2}{19.21}+\dfrac{-2}{21.23}+\dfrac{-2}{23.25}+-\dfrac{2}{25.27}-\dfrac{1}{27}\)Đặt:
\(A=\dfrac{-2}{1.3}+\dfrac{-2}{3.5}+\dfrac{-2}{5.7}+...+\dfrac{-2}{19.21}+\dfrac{-2}{21.23}+\dfrac{-2}{23.25}+\dfrac{-2}{25.27}\)\(A=-1\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{25}-\dfrac{1}{27}\right)\)
\(A=-1\left(1-\dfrac{1}{27}\right)\)
\(A=-1.\dfrac{26}{27}=-\dfrac{26}{27}\)
Thay vào biểu thức ban đầu ta có:
Giá trị là:
\(-\dfrac{26}{27}-\dfrac{1}{27}=-\dfrac{27}{27}=-1\)
BJ
4
C
6 tháng 4 2022
2/25=.2../25 .27../36=3/4
3/.12..=9/36
15/27=5/..9..
..20..28=5/7
36/72=..1../2
6 tháng 4 2022
\(\dfrac{2}{25}=\dfrac{2}{25}\)
\(\dfrac{27}{36}=\dfrac{3}{4}\)
\(\dfrac{3}{12}=\dfrac{9}{36}\)
\(\dfrac{15}{27}=\dfrac{5}{9}\)
\(\dfrac{20}{28}=\dfrac{5}{7}\)
\(\dfrac{36}{72}=\dfrac{1}{2}\)
MN
9 tháng 1 2016
10) 37.(27+25)-27.(37+25)=37.27+37.25-27.37-27.25
=37.25-27.25=25(37-27)=25.10=250
11)3.(-5+3)+7(2+5)=3.(-2)+7.7=-6+49=43
12)-3.(2-7)-6(8-5)=-3.(-5)-6.3=15-18=-3
T
10 tháng 9 2023
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
20 tháng 12 2021
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)