\(C=x^7-80x^6+80x^5-80x^4+80x^3-80x^2+80x+15\)

Ta có x=79 => 80=79+1=x+1

\(C=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x+15\)

\(C=x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x+15\)

\(C=x+15=79+15=94\)

Có : x = 79

=> x + 1 = 80

Xét P(x) , có :

\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+....+80x+15\)

\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+....+\left(x+1\right)x+15\)

\(P\left(x\right)=x^7-x^7-x^6+x^6+x^5-x^5-x^4+....+x^2+x+15\)

\(P\left(x\right)=x+15\)

\(P\left(79\right)=79+15=94\)

Thay x+1=80 ta đc:

\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)

\(=x^7-x^7-x^6+x^6+x^5+...+x^2+x+15\)

\(79+15=94\)

\(Ta \)  \(có \) \(:\)

\(x = 79 \)\(\Rightarrow\)\(x + 1 = 80\)

\(Thay \)  \(x + 1 = 80 \) \(vào \)  \(P(x)\) \(ta\) \(được :\)

\(P ( x ) = x ^7 - ( x + 1 )x ^6 + ( x + 1 )x^5\)\(- ( x + 1 )x ^4\)\(+ ...+ ( x + 1 )x + 15\)

\(P ( x ) = x ^7 - x ^7- x^6 + x^6 + x^5 - x^ 5\)\(- x ^4 + x ^4 + ... - x^ 2 + x ^2 + x + 15\)

\(P ( x ) = x + 15\)

\(Thay x = 79 vào P ( x ) ta được :\)

\(P ( x ) = 79 + 15 = 94\)

\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4\)\(+...+\left(x+1\right)x+15\)

\(P\left(x\right)=x^7-x^7-x^6+x^6+...+x^2+x+15\)

\(P\left(x\right)=x+15=94\)

Vậy giá trị của P(x) tại x = 79 là 94 

ng qua ko bt lam a bn

Nguyễn Khắc Nam m khủng hả con

P(x)=x7−80x6+80x5−8x4+...+80x+15

⇒P(x)=x7−(x+1).x6+(x+1).x5+...+(x+1)x+15

⇒P(x)=x7−x7−x6+x6+x5−x5+...−x3−x2+x2+x+15

⇒P(x)=x+15 (1)

Thay x=79 vào (1),ta được:

P(79)=79+15=84

~ Học tốt ~

x= 79hihi

x=79hihi

a)p(x)=x⁷-80x⁶+80x⁵-80x⁴+.........+80+15

=x⁷-(79+1)⁶+(79+1)⁵-(79+1)⁴+.........+(79+1)x+15

mà x=79

=> x⁷-(x+1)⁶+(x+1)⁵-(x+1)⁴+..........+(x+1)x+15

=x⁷-x⁷+x⁶-x⁶+x⁵-x⁵+........+x²+x+15

=x+15

=79+15

=94

\(P\left(x\right)=x^7-80x^6+80x^5-8x^4+...+80x+15\)

\(\Rightarrow P\left(x\right)=x^7-\left(x+1\right).x^6+\left(x+1\right).x^5+...+\left(x+1\right)x+15\)

\(\Rightarrow P\left(x\right)=x^7-x^7-x^6+x^6+x^5-x^5+...-x^3-x^2+x^2+x+15\)

\(\Rightarrow P\left(x\right)=x+15\) \(^{\left(1\right)}\)

Thay \(x=79\) vào \(^{\left(1\right)}\),ta được:

\(P\left(79\right)=79+15=84\)

b) Để \(\dfrac{n+3}{n-1}\) là số nguyên thì \(n+3⋮n-1\)

\(\Leftrightarrow4⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)