ai giúp mình với

10 cm á bn

10 m đúng hơn chứ

x + 20 ⋮ 5, x ϵ {15;17;50;23}

TH1: 

x = 15 ⇒ 15 + 20 = 35 ⇒ 35 ⋮ 5 ⇒ x = 15 thỏa mãn.

TH2:

x = 17 ⇒ 17 + 20 = 37 ⇒ 37 \(⋮̸\)5 ⇒ x = 17 không thỏa mãn.

TH3:

x = 50 ⇒ 50 + 20 = 70 ⇒ 70 ⋮ 5 ⇒ x = 50 thỏa mãn.

TH4:

x = 23 ⇒ 23 + 20 = 43 ⇒ 43 \(⋮̸\)5 ⇒ x = 23 không thỏa mãn.

⇒ x = 15;50 ⇒ x + 20 ⋮ 5

a) 3.5² + 15.2² - 26 : 2

= 3.25 + 15.4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) 2021 + 5.[300 - (17 - 7)²]

= 2021 + 5.[300 - 10²]

= 2021 + 5.[300 - 100]

= 2021 + 5.200

= 2021 + 1000

= 3021

c) 3².5 + 2³.10 - 81

= 9.5 + 8.10 - 81

= 45 + 80 - 81

= 125 - 81

= 44

a) 3.5² + 15.2² - 26 : 2

= 3. 25 + 15 . 4 - 26 : 2

=  75  + 60 - 13

= 135 - 13

= 122

b) 2021 + 5[300 - (17 - 7)² ]

= 2021 + 5[300 - 10² ]

=2021 + 5[300 - 100]

=2021 + 5.200

=2021 + 1000

= 3021

c) 3² . 5 + 2³ . 10 - 81

= 9 . 5 + 8 . 10 - 81

= 45 + 80 - 81

= 125 - 81

= 44

Google

A =  -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)

    = -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))

Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)

Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))

               = 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))

               = 2×_{\(\dfrac{24}{50}\)}

                   _{=  \(\dfrac{24}{25}\)}

      _{Thay B vào A ta có :}

   _{A = -1-\(\dfrac{24}{25}\)}

 _{=> A = \(\dfrac{-49}{25}\)}

 _{Cho mik một tick nhé thankss}

sai đề bài rồi bạn ơi

X=0

Y=1

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