(-1)+-39/13

a: \(\left(-2\right)^3-45:\left(-3\right)^2+\left(-2019\right)^0\cdot1^{2019}\)

\(=-8-45:9+1\)

\(=-8-5+1\)

=-13+1

=-12

b: \(11^{25}:11^{13}-3^5:\left(1^{10}+2^3\right)-60\)

\(=11^{25-13}-3^5:3^2-60\)

\(=11^{12}-27-60\)

\(=11^{12}-87\)

1125 : 3² + 4³ . 125 -125 : 5²

= (1125 : 9 ) + 64 . 125 - 125 : 25

= 125+ 64 . 125 - 125  : 25 

= 125 . ( 1 + 64 ) - 5

= 125 . 65 - 5 

=8125-5=8120

Bài 1: Thực hiện các phép tính sau:

\(a)\)Chưa rỏ đề

\(b)\)\(5025\div5-25\div5\)

\(=\)\(1005-5\)

\(=\)\(1000\)

\(c)\)\(218-180\div2\div9\)

\(=\)\(218-10\)

\(=\)\(208\)

\(d)\)\(\left(328-8\right)\div32\)

\(=\)\(320\div32\)

\(=\)\(10\)

Bài 1:

a) ( Tôi không nhìn rõ đầu bài )

b) 5025  :  5  -  25  :  5

=  ( 5025  -  25 )  :  5

=        5000   :   5

=       1000

c)   218  -  180  :  2  :  9

=     218   -    180  :  (  2  .  9 )

=      218    -    180   :  18

=       218    -    10

=       208

d)   (  328  -  8  )  :  32

=      320   :   32

=     10 

Bài 2: Thực hiện phép tính:

\(a)\)\(27.75+25.27-150\)

\(=\)\(27.\left(75+25\right)-150\)

\(=\)\(27.100-150\)

\(=\)\(2700-150\)

\(=\)\(2550\)

\(b)\)\(12\div\left\{400\div\left[500-\left(125+25.7\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-\left(125+175\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-300\right]\right\}\)

\(=\)\(12\div\left\{400\div200\right\}\)

\(=\)\(12\div2\)

\(=\)\(6\)

\(c)\)\(13.17-256\div16+14\div7-1\)

\(=\)\(221-16+2-1\)

\(=\)\(206\)

\(d)\)\(18\div3+182+3.\left(51\div17\right)\)

\(=\)\(6+182+3.3\)

\(=\)\(6+182+9\)

\(=\)\(197\)

\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)

\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)

\(=0,2-\dfrac{2}{3}\)

\(=-\dfrac{7}{15}\)

\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)

\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)

\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

#\(Toru\)

\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)