a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)

\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)

b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)

a)-1-2-3-4-5-6-....-80

=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)

Khoảng cách giữa các số:(-1)-(-2)=1

Tổng các số hạng:(-1)-(-80)+1=80 số

Tổng:[(-1)+(-80)].80:2= -3240

=>-1-2-3-4-5-6+......-80=-3240

b,1-2+3-4+5-6+......+2021-2022

=(1-2)+(3-4)+(5-6)+...+(2021-2022)

=(-1)+(-1)+(-1)+...+(-1)

Tổng số cặp là:

(2022-1+1):2=1011 cặp

-1.1011=-1011

=>1-2+3-4+5-6+......+2021-2022= -1011

c, Đề bài sai

d,-4-8-12-16-.......-2020

=-4+(-8)+(-12)+(-16)+...+(-2020)

Khoảng cách giữa các số:-4-(-8)=4

Tổng các số hạng:[-4-(-2020]:4+1=505 số

Tổng:[-4+(-2020)].505:2=-511060

=>-4-8-12-16-.......-2020=-511060

Mọi người giúp mình với ạ! Thank you everyone!

`@` `\text {Ans}`

`\downarrow`

`a.`

\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-0,4+1\)

`= -0,1 + 1`

`= 0,9`

`b.`

\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)

`=`\(1+4\cdot\left(-2,25\right)\)

`= 1+ (-9) = -8`

`c.`

\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)

`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)

`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)

`=`\(\dfrac{13}{6}\div2\)

`=`\(\dfrac{13}{12}\)

`d.`

\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)

`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)

`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)

`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)

`e.`

\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)

`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)

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

`f.`

\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)

`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)

`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)

`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)

`= 0,8 \div (-2/15)`

`=-6`

`@` `yHGiangg.`

a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)

\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)

\(=\dfrac{5}{7}\)

a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)

\(=8\cdot5^2+5-1\)

\(=200+4=204\)

b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)

\(=27-1+125-1\)

=150

a. (8,5⁷ - 5⁵ + 5⁶) : 5⁵

= (8,5⁷ : 5⁵) - (5⁵ : 5⁵) + (5⁶ : 5⁵)

= 1,7² - 1 + 5

= 2,89 - 1 + 5

= 6,89

b. (9³⁰ - 27¹⁹) : 3⁵⁷ + (125⁹ - 25¹²) : 5²⁴

= (9³⁰ : 3⁵⁷) - (27¹⁹ : 3⁵⁷) + (125⁹ : 5²⁴) - (25¹² : 5²⁴)

= 3³ - 1 + 125 - 1

= 27 - 1 + 125 - 1

= 150

c. (10¹² + 5¹¹ . 2⁹ - 5¹³ - 2⁸) : 4 . 5⁵ . 10⁶

= (10¹² + 2,5 , 10¹⁰ - 5¹³ - 2⁸) : 1,25 . 10¹⁰

= (10¹² : 1,25 . 10¹⁰) + (2,5 . 10¹⁰ : 1,25 . 10¹⁰) - (5¹³ : 1,25 . 10¹⁰) - (2⁸ : 1,25 . 10¹⁰)

= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)

= 81,90234373 \(\approx\) 82

a) \(\left( { - 3} \right).\left( { - 2} \right).\left( { - 5} \right).4\)\( = \left[ {\left( { - 3} \right).\left( { - 2} \right)} \right].\left( { - 5} \right).4\)\( = 6.\left( { - 5} \right).4 =  - 30.4 =  - 120\).

b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).

8-3-+7=12

-5-9+12=-2

7+9-3=13

-3+8-11=-6

8-3+7=12

(-5)-9+12=-2

7-9-3=-5

(-3)+8-11=-6

a, 8-(3-7)

=8-(-4)

=4

b,(-5)-(9-12)

=(-5)-(-3)

=(-8)

c,7-(-9)-3

=(-2)-3

=-1

d.(-3)+8-11

=5-11

=6

\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)

\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)

a) 49 - 9 . [ 3⁴ : 3¹ - ( 12 - 4 ) : 2²  ]

= 49 - 9 . [ 3³ - 8 : 2² ]

=  49 - 9 . [ 3³ -2³ : 2² ]

=  49 - 9 . [ 3³ - 2 ]

= 49 - 9 . [ 27 - 2 ]

= 49 - 9 . 25

= 49 - 225 = -176

b) 4.5² + 16 : 2²

= 4. 25 + 2⁴ : 2²

= 100 + 2² = 100+ 4 =104

--tho--là tớ💋

c) 5³ : 5 - 2017⁰

= 5² - 1 = 25 - 1= 24 

d) 2² . 3 - ( 10⁰ + 8 ) : 3²

= 4. 3 - ( 1 + 8 ) : 3²

= 4. 3 - 9 : 3²

= 12 - 3² : 3²

= 12 - 1 = 11.

--tho--là tớ💋