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

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

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

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

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

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

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

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

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

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

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}

Câu 1:

Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

=>OA+AB=OB

=>AB+4=10

=>AB=6(cm)

Vì BM<BA

nên M nằm giữa B và A

=>BM+MA=BA

=>MA+4=6

=>MA=2(cm)

Câu 2:

\(3xy+2x-5y=6\)

=>\(x\left(3y+2\right)-5y-\dfrac{10}{3}=\dfrac{8}{3}\)

=>\(3x\left(y+\dfrac{2}{3}\right)-5\left(y+\dfrac{2}{3}\right)=\dfrac{8}{3}\)

=>\(\left(3x-5\right)\left(3y+2\right)=8\)

=>\(\left(3x-5;3y+2\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(2;2\right);\left(\dfrac{13}{3};-\dfrac{1}{3}\right);\left(\dfrac{4}{3};-\dfrac{10}{3}\right);\left(-1;-1\right);\left(\dfrac{7}{3};\dfrac{2}{3}\right);\left(9;0\right);\left(1;-2\right);\left(\dfrac{1}{3};-\dfrac{4}{3}\right)\right\}\)

mà x,y nguyên

nên \(\left(x;y\right)\in\left\{\left(2;2\right);\left(-1;-1\right);\left(9;0\right);\left(1;-2\right)\right\}\)

\(8n-1⋮4n-5\)

\(\Rightarrow2\times4n-10+9⋮4n-5\)

\(\Rightarrow2\times4n-2\times5+9⋮4n-5\)

\(\Rightarrow2\left(4n-5\right)+9⋮4n-5\)

Vì \(2\left(4n-5\right)⋮4n-5\) 

\(\Rightarrow9⋮4n-5\)

\(\Rightarrow4n-5\) \(\inƯ\left(9\right)=\left\{1,3,9,-1,-3,-9\right\}\) 

\(\Rightarrow4n\in\left\{6;8;14;4;2;-4\right\}\)

mà \(4n⋮4\) 

\(\Rightarrow4n\in\left\{8;4;-4\right\}\)

\(\Rightarrow n\in\left\{2,1,-1\right\}\)