Bài 1 :

a) \(25.8^3-23.8^3\)

\(=8^3.\left(25-23\right)\)

\(=512.2=1024\)

b) \(5^4-2.5^3\)

\(=5^3.5-2.5^3\)

\(=5^3\left(5-2\right)\)

\(=125.3\)

\(=375\)

c) \(2.4^3-4^3.7-6.4^3\)

\(=4^3.\left(2-7-6\right)\)

\(=64.\left(-11\right)=-704\)

d) \(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right]\)

\(=9.1000-\left[169-\left(25.4+4.15\right)\right]\)

\(=9000-\left[169-4\left(25+15\right)\right]\)

\(=9000-\left[169-4.40\right]\)

\(=9000-\left[169-160\right]\)

\(=9000-9=8991\)

a.\(25.8^3-23.8^3=8^3.\left(25-23\right)=512.2=1024\)

b. \(5^4-2.5^3=5^3.\left(5-2\right)=125.3=375\)

c. \(2.4^3-4^3.7-6.4^3=4^3.\left(2-7-6\right)\)\(=64.\left(-11\right)=-704\)

d. \(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right]=\)\(3^2.10^3-\left[13^2-\left(100+60\right)\right]\) \(=3^2.10^3-\left(13^2-160\right)=9000-9=8991\)

a: \(2^2\cdot5-\dfrac{\left(1^{10}+8\right)}{3^2}\)

\(=4\cdot5-\dfrac{1+8}{3}\)

=20-3

=17

b: \(5^8:5^6+4\left(3^2-1\right)\)

\(=5^2+4\left(9-1\right)\)

=25+4*8

=25+32

=57

c: \(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)

\(=400-36+20:\left[27-5\right]\)

\(=364+\dfrac{20}{22}\)

\(=364+\dfrac{10}{11}=\dfrac{4014}{11}\)

A) 2².5 - (1¹⁰ + 8) : 3²

= 4.5 - (1 + 8) : 9

= 20 - 9 : 9

= 20 - 1

= 19

B) 5⁸ : 5⁶ + 4.(3² - 1)

= 5² + 4.(9 - 1)

= 25 + 4.8

= 25 + 32

= 57

C) 400 - {36 - 20 : [3³ - (8 - 3)]}

= 400 - [36 - 20 : (27 - 5)]

= 400 -(36 - 20 : 22)

= 400 - (36 - 10/11)

= 400 - 386/11

= 4014/11

\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)

\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)

a,32

b,\(-\frac{1}{10}\)

c,-1000000

d,\(\frac{9}{16}\)

a,

\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)

b,

\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)

c,

\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

d,

\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)