(2018x+3y+1)(2018x+2018x+y) =225
Tìm x, y, biết: \(\left(2018x+3y+1\right).\left(2018^x+2018x+y\right)=225\)
F(x)=x7-2018x6+2018x5-2018x4+2018x3-2018x2+2018x+1 Với x=2017
F(x)=\(x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x+1.\)
x=2017=>2018=x+1 thay vào F(x) ta có:
F(x)=x+1=2018
tinh p=x\(^{15}\)-2018x\(^{14}\)+2018x\(^{13}\)-2018x\(^{12}\)+...+2018x\(^3\)-2018x\(^2\)+2018x-2018 ;voi x=2017
Ta có: x=2017
nên x+1=2018
Ta có: \(P=x^{15}-2018x^{14}+2018x^{13}-2018x^{12}+...+2018x^3-2018x^2+2018x-2018\)
\(=x^{15}-\left(x+1\right)\cdot x^{14}+\left(x+1\right)\cdot x^{13}-\left(x+1\right)\cdot x^{12}+...+\left(x+1\right)\cdot x^3-\left(x+1\right)\cdot x^2+\left(x+1\right)\cdot x-\left(x+1\right)\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}+...+x^3-x^3+x^2-x^2+x-x-1\)
=-1
Cho x = 2017. Tính giá trị của đa thức
\(P\left(x\right)=x^{2017}-2018x^{2017}+2018x^{2016}-2018x^{2015}+...+2018x^2-2018x+1\)
\(P\left(x\right)=x^{2017}-2018x^{2017}+2018x^{2016}-...-2018x+1\)
Vì \(x=2017\)
\(\Leftrightarrow x+1=2018\)
Thay vào P(x) ta được :
\(P\left(x\right)=x^{2017}-x^{2017}\left(x+1\right)+x^{2016}\left(x+1\right)-...-x\left(x+1\right)+1\)
\(P\left(x\right)=x^{2017}-x^{2018}-x^{2017}+x^{2017}+x^{2016}-...-x^2-x+1\)
\(P\left(x\right)=-x^{2018}+1\)
\(P\left(x\right)=-2017^{2018}+1\)
Cho x \(=\)2017, Tính giá trị biểu thức
A\(=\)\(x^9-2018x^8+2018x^7-2018x^6+2018x^5-2018x^4-2018x^3-2018x^2+2018x-2018\)
\(A=x^9-2018x^8+2018x^7-2018x^6+2016x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)
\(A=x^9-\left(2017+1\right)x^8+\left(2017+1\right)x^7-...+\left(2017+1\right)x-\left(2017+1\right)\)
\(A=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-...+\left(x+1\right)x-x-1\)
\(A=x^9-x^9-x^8+x^8+x^7-...+x^2+x-x-1\)
\(A=-1\)
Mk sửa lại đề. bn tham khảo nha!!!
\(x=2017\)\(\Rightarrow\)\(x+1=2018\)
Ta có: \(A=x^9-2018x^8+2018x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)
\(=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\)
\(=x^9-x^9-x^8+x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-1\)
\(=1\)
Tính P (2017) biết rằng:
P (x)=x^2017- 2018x^2016+2018x^2015-2018x^2014+...-2018x^2+2018x-1
GIÚP MÌNH VỚI!!!!PLEASE!!!!😖😖😖
P(x)= x^2017 - 2018x^2016+ 2018x^2015+...-2018x^2 + 2018x-1
=> P(x)= x^2017 -2017x^2016-x^2016 + 2017x^2015 + x^2015+..-2017x^2-x^2 + 2017x+x-1
=> P(x)= x^2016(x-2017) -x^2015(x-2017)+...- x(x -2017)+ x-1
thay x=2017 vào p(x) ta được
p(2017)= 2016
tính giá trị của biểu thức B=2018x^100 + 2018x^99 + 2018x^98 + ... + 2018x^2 + 2018x . Tại x = 2017
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mk tick cho
Tính \(A=x^5-2018x^4+2018x^3-2018x^2+2018x-1000\) tại x=2017
Lời giải:
Ta có:
\(A=x^5-2018x^4+2018x^3-2018x^2+2018x-1000\)
\(A=(x^5-2017x^4)-(x^4-2017x^3)+(x^3-2017x^2)-(x^2-2017x)+x-1000\)
\(A=x^4(x-2017)-x^3(x-2017)+x^2(x-2017)-x(x-2017)+x-1000\)
Tại \(x=2017\Rightarrow A=2017^4.0-2017^3.0+2017^2.0-2017.0+2017-1000\)
\(A=2017-1000=1017\)
tính đi đã thức hay đơn thức ý các bạn hộ mình với ạ!!!mơn@@
F=x^11-2018x^10-2018x^9-2018x^8-2018x^7-...-2018x-2017 với x=2105 ( hình như x=2017 hay sao ý ạ)