Phép nhân và phép chia các đa thức

8

8

n^3 – n +6n^2 -18n -24
=n(n^2 -1 ) +6 (n^2- 3n -4)
=n(n-1)(n+1) +6(n-4)(n+1)
vì n(n-1)(n+1) chia hết cho 2,3 => chia hết cho 6.
6(n-4)(n+1) cũng chia hết cho 6
=> A chia hết cho 6

đặt x+29=y

ta có :

\(\dfrac{1}{y^2}+\dfrac{1}{\left(y+1\right)^2}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{2y^2+2y+1}{y^4+2y^3+y^2}=\dfrac{5}{4}\)

\(\Leftrightarrow5y^4+10y^3-3y^2-8y-4=0\)

\(\Leftrightarrow\left(y-1\right)\left(5y^3+15y^2+12y+4\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(y+2\right)\left(5y^2+5y+2\right)=0\)

\(\left[{}\begin{matrix}y-1=0\Leftrightarrow y=1\Rightarrow x+29=1\Leftrightarrow x=-28\\y+2=0\Leftrightarrow y=-2\Rightarrow x+29=-2\Leftrightarrow x=-31\\5y^2+5y+2=0\left(loại\right)\end{matrix}\right.\)

vậy, S={-31;-28}

Xài delta nhé!

Cho Q(x) = 0 hay \(3x^2+5x-21=0\)

Ta có:

\(\Delta=b^2-4ac=5^2-4.3.\left(-21\right)=277\)

Vì \(\Delta>0\) nên phương trình có 2 nghiệm phân biệt.

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-5+\sqrt{277}}{6}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-\sqrt{277}}{6}\)

Vậy..............

Ta có: \(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\) < A
Vậy A > B

Ta có:

\(A=2^{16}\)

\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^8-1\right)\left(2^8+1\right)\)

\(B=2^{16}-1< 2^{16}\)

Vậy A > B

Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)

Vì \(2^{16}>2^{16}-1\Rightarrow A>B\)

1001

<=> (2⁴-1)(2⁴+1).....(2^m+1)=2^3m-218

^<=> 2^2m-1+1=2^3m-218

^<=> 2^2m=2^3m-218

<=>2m=3m-218

=>m=218

<=> (24-1)(24+1).....(2m+1)+1=23m-218

<=> 22m-1+1=23m-218

<=> 22m=23m-218

<=>2m=3m-218

=>m=218

ở dưới mình nhầm nha!!!