TÍNH GIÁ TRỊ CỦA ĐA THỨC
P= X^7-80x^6+80x^5-80x^4+...+80x+15 v ới x=79
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\(C=x^7-80x^6+80x^5-80x^4+80x^3-80x^2+80x+15\)
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:
\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)
\(=x^7-x^7-x^6+x^6+x^5+...+x^2+x+15\)
\(79+15=94\)
\(Ta \) \(có \) \(:\)
\(x = 79 \)\(\Rightarrow\)\(x + 1 = 80\)
\(Thay \) \(x + 1 = 80 \) \(vào \) \(P(x)\) \(ta\) \(được :\)
\(P ( x ) = x ^7 - ( x + 1 )x ^6 + ( x + 1 )x^5\)\(- ( x + 1 )x ^4\)\(+ ...+ ( x + 1 )x + 15\)
\(P ( x ) = x ^7 - x ^7- x^6 + x^6 + x^5 - x^ 5\)\(- x ^4 + x ^4 + ... - x^ 2 + x ^2 + x + 15\)
\(P ( x ) = x + 15\)
\(Thay x = 79 vào P ( x ) ta được :\)
\(P ( x ) = 79 + 15 = 94\)
\(C=x^7-80x^6+80x^5-80x^4+80x^3-80x^2+80x+15\)
\(=x^7-79x^6-x^6+79x^5+x^5-79x^4-x^4+79x^3+x^3-79x^2-x^2+79x+x-79+94\)
\(=x^6\left(x-79\right)-x^5\left(x-79\right)+x^4\left(x-79\right)-x^3\left(x-79\right)+x^2\left(x-79\right)-x\left(x-79\right)+\left(x-79\right)+94\)
\(=\left(x^6-x^5+x^4-x^3+x^2-x+1\right)\left(x-79\right)+94\)
Thay x = 79 \(\Rightarrow C=94\)
Vậy C = 94 khi x = 79
Thay x = 79 vào C ta có:
C =\(79^7-80.79^6+80.79^5-80.79^4+80.79^3-80.79^2+80.79+15\)
C = \(79^7-\left(79+1\right).79^6+\left(79+1\right).79^5-\left(79+1\right).79^4+\left(79+1\right).79^3-\left(79+1\right).79^2+\left(79+1\right).79+15\)
C = \(79^7-79^7+79^6-79^6+79^5-79^5+79^4-79^4+79^3-79^3+79^2-79^2+79+15\)
C = 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
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 ~
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:
\(P\left(79\right)=79+15=84\)
Vì x = 79
=> x + 1 = 80
Khi đó P = x7 - 80x6 + 80x5 - 80x4 + ... + 80x + 15
= x7 - (x + 1)x6 + (x + 1)x5 - (x + 1)x4 + ... + (x + 1)x + 15
= x7 - x7 + x6 + x6 + x5 - x5 - x4 + ... + x2 + x + 15
= x + 15 = 79 + 15 = 94
Vậy P = 94 khi x = 15