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22 tháng 4 2017

\(\left(x+2,5\right)\) \(:2\dfrac{1}{2}\) \(-1,5\) \(=-1\dfrac{1}{4}\)

\(\left(x+\dfrac{5}{2}\right):\dfrac{5}{2}-\dfrac{3}{2}=-\dfrac{5}{4}\)

\(\left(x+\dfrac{5}{2}\right):\dfrac{5}{2}=\dfrac{1}{4}\)

\(x+\dfrac{5}{2}=\dfrac{5}{8}\)

\(\Rightarrow x=-\dfrac{15}{8}\)

Vậy \(x=-\dfrac{15}{8}\)

`@` `\text {Ans}`

`\downarrow`

`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`

`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`

`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`

`=> x^2 . 11 = 2^4 . 11`

`=> x^2 . 11 - 2^4 . 11 = 0`

`=> 11 . (x^2 - 16) = 0`

`=> x^2 - 16 = 0`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = `\(\pm4\)

Vậy, `x \in`\(\left\{4;-4\right\}\)

_____

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)

`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)

`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)

`=>`\(\dfrac{23}{108}x=48-6^2\)

`=>`\(\dfrac{23}{108}x=48-36\)

`=>`\(\dfrac{23}{108}x=12\)

`=>`\(x=\dfrac{1296}{23}\)

Vậy, `x = `\(\dfrac{1296}{23}\)

13 tháng 7 2023

\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)

\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)

\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)

\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)

\(\Rightarrow x\in\varnothing\)

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)

\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)

11 tháng 5 2023

1)\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{11}{70}\)

\(\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{x\left(x+3\right)}\right):3=\dfrac{11}{70}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{11}{70}\cdot3\)

\(\dfrac{1}{2}-\dfrac{1}{x+3}=\dfrac{33}{70}\)

\(\dfrac{1}{x+3}=\dfrac{1}{2}-\dfrac{33}{70}\)

\(\dfrac{1}{x+3}=\dfrac{2}{70}\)

\(\dfrac{1}{x+3}=\dfrac{1}{35}\)

\(x+3=35\\ x=35-3\\ x=32\)

2) Số góc đc tạo thành từ 2023 tia chung gốc là:\(\dfrac{2023\cdot2022}{2}=2045253\) (góc)

11 tháng 5 2023

Bài 1 thì bạn Ánh làm đúng rồi

Bài 2 thì giải chi tiết như này em nhé:

Cứ 1 tia tạo với 2023 - 1 tia còn lại là 2023 - 1 góc

Với 2023 tia thì tạo được số góc là:  (2023 - 1)\(\times\) 2023 góc

Theo cách tính trên thì mỗi góc đã được tính hai lần

Vậy số góc tạo được là: (2023-1)\(\times\) 2023: 2 = 2045253 (góc)

Kết luận: ...

4 tháng 4 2022

\(x\) \((\)\(\dfrac{3}{2.5}\) \(+ \) \(\dfrac{3}{5.8}\) \(+\) \(\dfrac{3}{8.11}\) \(+\) \(\dfrac{3}{11.14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{8}\) \(+\) \(\dfrac{1}{8}\) \(-\) \(\dfrac{1}{11}\) \(+\) \(\dfrac{1}{11}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) x \(\dfrac{3}{7}\) \(=\) \(\dfrac{1}{21}\)
\(x\)        \(=\) \(\dfrac{1}{21}\) \(:\) \(\dfrac{3}{7}\) 
\(x\)        \(=\) \(\dfrac{1}{9}\)

 

Câu 2: 

\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)

\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)

\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)

26 tháng 7 2021

1,\(\dfrac{-1}{4}-\dfrac{3}{4}:x=-\dfrac{11}{36}\)

\(-\dfrac{3}{4}:x=\left(-\dfrac{1}{4}\right)-\left(-\dfrac{11}{36}\right)\)

\(-\dfrac{3}{4}:x=\dfrac{1}{18}\)

\(x=\left(-\dfrac{3}{4}\right):\left(\dfrac{1}{18}\right)\)

\(x=\dfrac{27}{2}\)

2, \(\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\)

\(\dfrac{3}{4}x=\dfrac{3}{7}+\dfrac{1}{2}\)

\(\dfrac{3}{4}x=\dfrac{13}{14}\)

\(x=\dfrac{13}{14}:\dfrac{3}{4}\)

\(x=\dfrac{26}{21}\)

26 tháng 7 2021

1 , x = 36/11

2 , x = 26/21

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

\(x=\dfrac{6}{5}-\dfrac{2}{3}\)

\(x=\dfrac{18}{15}-\dfrac{10}{15}\)

\(x=\dfrac{8}{15}\)

Vậy, `x =`\(\dfrac{8}{15}\)

`b)`

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{4}{17}\)

Vậy, \(x=\dfrac{4}{17}\)

`c)`

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{34}{7}\)

Vậy, `x = `\(\dfrac{34}{7}\)

13 tháng 7 2023

a) \(\dfrac{3}{2}x\dfrac{4}{5}-x=\dfrac{2}{3}\Rightarrow\dfrac{6}{5}-x=\dfrac{2}{3}\Rightarrow x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18}{15}-\dfrac{10}{15}=\dfrac{8}{15}\)

b) \(x.3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}:\dfrac{17}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}.\dfrac{4}{17}\Rightarrow x=\dfrac{10}{3}.\dfrac{4}{17}:\dfrac{10}{3}=\dfrac{10}{3}.\dfrac{4}{17}.\dfrac{3}{10}=\dfrac{4}{17}\)

c) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\dfrac{1}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{11}{3}-\dfrac{5}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{22}{6}-\dfrac{15}{6}\Rightarrow\dfrac{17}{3}:x=\dfrac{7}{6}\Rightarrow x=\dfrac{17}{3}:\dfrac{7}{6}=\dfrac{17}{3}.\dfrac{7}{6}=\dfrac{119}{18}\)

14 tháng 3 2022

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23 tháng 9 2023

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\)\(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

 

 

 

 

23 tháng 9 2023

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)

\(=\dfrac{1}{2016}\)

b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)

\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)

hay x=10

Vậy: x=10