Lấy 28 viên bi nkaa

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)

\(\Rightarrow3\cdot\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)

\(\Rightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}:3\)

\(\Rightarrow \left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{27}\)

\(\Rightarrow\left(3x-\dfrac{1}{2}\right)^3=\left(-\dfrac{1}{3}\right)^3\)

\(\Rightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Rightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}\)

\(\Rightarrow3x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}:3=\dfrac{1}{18}\)

\(x+\dfrac{13}{15}=-\dfrac{2}{5}\\ x=-\dfrac{2}{5}-\dfrac{13}{15}\\ x=-\dfrac{6}{15}-\dfrac{13}{15}\\ x=-\dfrac{19}{15}\)

Ta thấy :

\(\dfrac{3}{10}>\dfrac{3}{15}\)

\(\dfrac{3}{11}>\dfrac{3}{15}\)

\(\dfrac{3}{12}>\dfrac{3}{15}\)

\(\dfrac{3}{13}>\dfrac{3}{15}\)

\(\dfrac{3}{14}>\dfrac{3}{15}\)

\(\Rightarrow S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>5.\dfrac{3}{15}\) (1)

\(S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{1}{14}>1\)

Ta có:

\(\dfrac{3}{10}< \dfrac{3}{9}\)

\(\dfrac{3}{11}< \dfrac{3}{9}\)

\(\dfrac{3}{12}< \dfrac{3}{9}\)

\(\dfrac{3}{13}< \dfrac{3}{9}\)

\(\dfrac{3}{14}< \dfrac{3}{9}\)

\(\Rightarrow S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< 5.\dfrac{3}{9}\)

\(\Rightarrow S< \dfrac{5}{3}< 2\)

giải s ta có s=1.267782218 thì lớn hơn 1 và nhỏ hơn 2 

cíu mik vs mik đang cần gấp

  \(-\dfrac{1}{12}\) < \(\dfrac{x}{24}\) < \(\dfrac{1}{8}\)

  - \(\dfrac{2}{24}\) < \(\dfrac{x}{24}\) < \(\dfrac{3}{24}\)

- 2 < \(x\) < 3

  \(x\) \(\in\) {-1; 0; 1; 2}

Vậy \(x\) \(\in\) {-1; 0; 1; 2}

35%\(x\) + 0,5\(x\) + \(x\): 2 = 27

   \(x\) \(\times\) 0,35 + \(x\) \(\times\) 0,5 + \(x\) \(\times\) 0,5 = 27

  \(x\) \(\times\) (0,35 + 0,5 + 0,5) = 27

  \(x\) \(\times\) 1,35 = 27

 \(x\)              = 27 : 1,35

\(x\)               = 20

Vậy \(x\) = 20

Đúng0

giúp mik đi! ai nhanh đúng mik tik cho

ok

\(M=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

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

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)

ta có n^2 +5 =n(n+1)-(n+1)+6chia hết cho n+1

mà n(n+1)chia hết cho n+1; n+1chia hết cho n+1 

=> 6 chia hết cho n+1

=> n+1 thuộc ước của 6={1;-1;6;-6}

 =>n thuộc {0;-2;5;-7}

Vậy n thuộc {0;-2;5;-7}

ta có n^2 +5 =n(n+1)-(n+1)+6chia hết cho n+1

mà n(n+1)chia hết cho n+1; n+1chia hết cho n+1 

=> 6 chia hết cho n+1

=> n+1 thuộc ước của 6={1;-1;6;-6}

 =>n thuộc {0;-2;5;-7}

Vậy n thuộc {0;-2;5;-7}