b: \(35-2\cdot1^{111}+3\cdot7\cdot7^2\)

\(=35-2+21\cdot7^2\)

=33+147

=180

\(a,\) Số số hạng là \(\left(40-2\right):2+1=20\left(số\right)\)

Tổng là \(\left(40+2\right)\times20:2=420\)

\(b,\) Số số hạng là \(\left(39-1\right):2+1=20\left(số\right)\)

Tổng là \(\left(39+1\right)\times20:2=400\)

a) 2 + 4 + 6 + 8 + ... + 34 + 36 + 38 + 40

= ( 2 + 42 ) + ( 4 + 38 ) + .... + ( 20 + 22 )

= 42 \(\times\) 10

= 420

b) 1 + 3 + 5 + 7 + ... + 35 + 37 + 39

= ( 1 + 39 ) + ( 3 + 37 ) + ...+ ( 19 + 21 )

= 40 \(\times\) 10

= 400

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

Bài 1: 

a: \(-2^{30}=-8^{10}\)

\(-3^{30}=-27^{10}\)

mà 8<27

nên \(-2^{30}>-3^{30}\)

b: \(35^5=35^5\)

\(6^{10}=36^5\)

mà 35<36

nên \(35^5< 6^{10}\)

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

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

a) \(\dfrac{1}{3}+\dfrac{4}{3}\times\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{4}{6}=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

b) \(\dfrac{3}{5}\times\dfrac{4}{7}:\dfrac{16}{21}=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{21}{16}=\dfrac{12}{35}\times\dfrac{21}{16}=\dfrac{252}{560}=\dfrac{9}{20}\)

Bài 1: 

\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

\(\Rightarrow-x=1-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

\(x+4=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}-4\)

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

\(x-\dfrac{1}{5}=2\)

\(\Rightarrow x=2+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

a: \(3^8:3^4+2^2\cdot2^3\)

=81+32

=123

b: \(3\cdot4^2-2\cdot3^2\)

\(=48-18\)

=30

a, 3⁸: 3⁴+ 2². 2³

= 3^8-4 + 2²⁺³

= 3⁴ + 2⁵

= 81 + 32

= 113

b,  3 . 4²- 2 . 3²

= 3 . 16 - 2 . 9

= 48 - 18

= 30

c,  84 : 4 + 3⁹: 3⁷+ 5⁰

= 84 : 4 + 3² + 1

= 84 : 4 + 9 + 1

= 21 + 9 + 1

= 31

d,  295 - ( 31 - 2² . 5)²

= 295 - ( 31 - 4 . 5 )²

= 295 - ( 31 - 20 )²

= 295 - 11²

= 295 - 121

= 174

e, 500 - {5[409 - (2³ . 3 - 21)² ] + 10³ } : 15

= 500 -  {5[409 - (8 . 3 - 21)² ] + 10³ } : 15

= 500 -  {5[409 - (24 - 21)² ] + 10³ } : 15

= 500 -  {5[409 - 3² ]+ 10³ } : 15

= 500 - {5[409 - 9 ]+ 10³ } : 15

= 500 - {5 . 400 + 1000 } : 15

= 500 - {2000 + 1000} : 15

= 500 - 3000 : 15

= 500 - 200

= 300

g, 5³ . 2 - 100 : 4 + 2³ . 5

= 125 . 2 - 100 : 4 + 8 . 5

= 250 - 25 + 40

= 225 + 40

= 265

h, 205 - [1200 - (4² - 2 . 3)³ ] : 40

= 205 - [ 1200 - ( 16 - 2 . 3 )³ : 40

= 205 - [ 1200 - ( 16 - 6 )³  ] : 40

= 205 - [ 1200 - 10³ ] : 40

= 205 - [ 1200 - 1000 ] : 40

= 205 - 200 : 40

= 205 - 5

= 200

Đây nha bạn!!!

\(a)2^3\cdot15-[115-(12-5)^2]\\=8\cdot15-(115-7^2)\\=120-(115-49)\\=120-66\\=54\\---\\b)5\cdot[(85-35:7):8+90]-60\\=5\cdot[(85-5):8+90]-60\\=5\cdot(80:8+90)-60\\=5\cdot(10+90)-60\\=5\cdot100-60\\=500-60\\=440\)

\(---\)

\(c,\left\{\left[261-\left(36-31\right)^3\cdot2\right]-9\right\}\cdot1001\)

\(=\left[\left(261-5^3\cdot2\right)-9\right]\cdot1001\)

\(=\left[\left(261-125\cdot2\right)-9\right]\cdot1001\)

\(=\left[\left(261-250\right)-9\right]\cdot1001\)

\(=\left(11-9\right)\cdot1001\)

\(=2\cdot1001\)

\(=2002\)

\(---\)

\(d,3\cdot10^2-\left[1200-\left(4^2-2\cdot3\right)^3\right]\)

\(=3\cdot100-\left[1200-\left(16-6\right)^3\right]\)

\(=300-\left(1200-10^3\right)\)

\(=300-\left(1200-1000\right)\)

\(=300-200\)

\(=100\)

#\(Toru\)

a) 2³.15 -[115-(12-5)² ]

= 2³.15 -[115-36]

= 8.15 -79

= 120-79

=41

b)5.[(85 - 35 : 7) :8 + 90 ] - 50

=5 .[80:8+90]-50

=5.100-50

=500-50

=450

c){[261 - ( 36-31)³.2 ]-9}.1001

={[261 - 125.2 ]-9}.1001

={[261 -250 ]-9}.1001

={11-9}.1001

=2.1001

=2002

d)3.10² - [1200 - ( 4² - 2.3)^3]

=3.100-[1200 -(16-6)³ ]

=300-[1200-1000]

=300-200

=100

`A=sqrt{(5-sqrt3)^2}+sqrt{(2-sqrt3)^2}`

`=5-sqrt3+2-sqrt3`

`=7-2sqrt3`

`B=sqrt{(3-sqrt2)^2}-sqrt{(1-sqrt2)^2}`

`=3-sqrt2-(sqrt2-1)`

`=4-2sqrt2`

`C=sqrt{(3+sqrt7)^2}-sqrt{(2-sqrt7)^2}`

`=3+sqrt7-(sqrt7-2)`

`=5`

`D=sqrt{4-2sqrt3}+sqrt{7+4sqrt3}`

`=sqrt{3-2sqrt3+1}+sqrt{4+2.2.sqrt3+3}`

`=sqrt{(sqrt3-1)^2}+sqrt{(2+sqrt3)^2}`

`=sqrt3-1+2+sqrt3=1+2sqrt3`

\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)

\(B=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)

\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)

\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|+\left|2+\sqrt{3}\right|\)

\(=\sqrt{3}-1+2+\sqrt{3}=1+2\sqrt{3}\)