a, 4⁸.8⁴

= (2²)⁸.(2³)⁴

= 2¹⁶.2¹²

= 2²⁸

b, 4¹⁵.5¹⁵

= (4.5)¹⁵

= 20¹⁵

c, 2¹⁰.15 + 2¹⁰.85

= 2¹⁰.(15 + 85)

= 2¹⁰.100

=2¹⁰.(2.5)²

= 2¹².5²

d, 3³.9²

= 3³ . (3²)²

= 3³.3⁴

= 3⁷

e, 5¹².7 - 5¹¹.10

= 5¹¹.(5.7 - 10)

= 5¹¹.25

=5¹¹.5²

=5¹³

f, \(x^1\).\(x^2\).\(x^3\)....\(x^{100}\)

= \(x^{1+2+3+...+100}\)

= \(x^{\left(1+100\right).100:2}\)

= \(x^{5050}\)

❤Em cảm ơn cô Nguyễn Thương hoài rất nhiều a!❤

a, 48.84

= (22)8.(23)4

= 216.212

= 228

b, 415.515

= (4.5)15

= 2015

c, 210.15 + 210.85

= 210.(15 + 85)

= 210.100

=210.(2.5)2

= 212.52

d, 33.92

= 33 . (32)2

= 33.34

= 37

e, 512.7 - 511.10

= 511.(5.7 - 10)

= 511.25

=511.52

=513

f, �1x1.�2x2.�3x3....�100x100

= �1+2+3+...+100x1+2+3+...+100

= �(1+100).100:2x(1+100).100:2

= �5050x5050

1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)

\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)

2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)

\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)

3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)

\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)

1.   \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

a: \(=\dfrac{-8}{9}-\dfrac{6}{5}+\dfrac{8}{9}=-\dfrac{6}{5}\)

c: \(=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

a, \(=\dfrac{1+4}{5}+\dfrac{5+1+3}{9}=1+1=2\)

b, \(=\dfrac{1+4+2}{3}+\dfrac{1+2+5}{6}=\dfrac{6}{3}+\dfrac{8}{6}=2+\dfrac{4}{3}=\dfrac{6+4}{3}=\dfrac{10}{3}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`-9/34*17/4`

`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)

`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)

`=`\(-\dfrac{9}{8}\)

`b)`

\(\dfrac{17}{15}\div\dfrac{4}{3}\)

`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)

`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)

`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)

`=`\(\dfrac{17}{20}\)

`c)`

\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)

`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)

`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)

`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)

a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)

\(=\dfrac{-9\cdot17}{34\cdot4}\)

\(=-\dfrac{153}{136}\)

\(=\dfrac{9}{8}\)

b) \(\dfrac{17}{15}:\dfrac{4}{3}\)

\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)

\(=\dfrac{17\cdot3}{15\cdot4}\)

\(=\dfrac{51}{60}=\dfrac{17}{20}\)

c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)

\(=\dfrac{21}{5}:-\dfrac{14}{5}\)

\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)

\(=\dfrac{21\cdot-5}{5\cdot14}\)

\(=-\dfrac{105}{70}=\dfrac{3}{2}\)

a) −934⋅17434−9​⋅417​

=−9⋅1734⋅4=34⋅4−9⋅17​

=−153136=−136153​

=98=89​

b) 1715:431517​:34​

=1715⋅34=1517​⋅43​

=17⋅315⋅4=15⋅417⋅3​

=5160=1720=6051​=2017​

c) 415:(−245)451​:(−254​)

=215:−145=521​:−514​

=215⋅−514=521​⋅−145​

=21⋅−55⋅14=5⋅1421⋅−5​

=−10570=32=−70105​=23​
 

a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)

b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)

\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)

\(=\dfrac{20}{720}=\dfrac{1}{36}\)