\(x^2+2x+4^n-2^{n+1}+2=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(4^n-2^n.2+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(2^n-1\right)^2=0\Rightarrow\orbr{\begin{cases}x=-1\\n=0\end{cases}}\)

\(C^n_n+C^{n-1}_n+C^{n-2}_n=37\)

\(\Leftrightarrow1+\dfrac{n!}{\left(n-1\right)!}+\dfrac{n!}{\left(n-2\right)!2!}=37\)

\(\Leftrightarrow1+n+\dfrac{n\left(n-1\right)}{2}=37\)

\(\Rightarrow n=8\)

\(P=\left(2+5x\right)\left(1-\dfrac{x}{2}\right)^8=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{x}{2}\right)^k\right)\)

\(=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)

\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5x\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)

\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\)

Số hạng chứa \(x^3\) trong \(2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\) là \(2C^3_8.\left(-\dfrac{1}{2}\right)^3x^3\)

Số hạng chứa \(x^3\) trong \(5\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\) là \(5C^2_8.\left(-\dfrac{1}{2}\right)^2x^3\)

Vậy số hạng chứa x³ trong P là:\(\left[2.C^3_8\left(-\dfrac{1}{2}\right)^3+5C^2_8\left(-\dfrac{1}{2}\right)^2\right]x^3\)

n^2+3n+2

=n^2+n+2n+2

=n(n+1)+2(n+1)

=(n+1)(n+2) chia hết cho n+1

a,18 chia hết cho n

=>n\(\in\)Ư(18)={-18,-9,-6,-3,-2,-1,1,2,3,6,9,18}

bai toan nay ?

Giải :

Câu 1 : Vì 18 ⋮ n nên n ∈ Ư ( 18 ) 

Ư ( 18 ) = { + 1 ; + 2 ; + 3 ; + 6 ; + 9 ; + 18 }

⇒ n ∈ { + 1 ; + 2 ; + 3 ; + 6 ; + 9 ; + 18 }

Câu 2 :  n + 2 ∈ Ư ( 16 ) 

Ư ( 16 ) = { + 1 ; + 2 ; + 4 ; + 8 ; + 16 }

⇒ n + 2 ∈ { + 1 ; + 2 ; + 4 ; + 8 ; + 16 }

⇒ n ∈ { - 3 ; - 1 ; - 4 ; 0 ; - 6 ; 2 ; - 10 ; 6 ; - 18 ; 14 }

Câu 3 : n - 4 ⋮ n - 1 ⇒ ( n - 1 ) - 3 ⋮ n - 1

Vì n - 1 ⋮ n - 1 , để ( n - 1 ) - 3 ⋮ n - 1 khi 3 ⋮ n - 1 ⇒ n - 1 ∈ Ư ( 3 ) = { + 1 ; + 3 }

 ⇒ n ∈ { 0 ; 2 ; - 2 ; 4 }

Các câu khác làm tương tự 

2^n+1= 32

=> 2^5=32

=> n=5-1

n=4

a)2ⁿ⁺¹=32

=>2ⁿ⁺¹=2⁵

=>n+1=5=>n=4

b)2ⁿ⁺²-2ⁿ⁼

=>2ⁿ(2²-1)=

=>2^n.5=

=>2ⁿ=

=>n=

=