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\(A=x^9-2018x^8+2018x^7-2018x^6+2016x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)
\(A=x^9-\left(2017+1\right)x^8+\left(2017+1\right)x^7-...+\left(2017+1\right)x-\left(2017+1\right)\)
\(A=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-...+\left(x+1\right)x-x-1\)
\(A=x^9-x^9-x^8+x^8+x^7-...+x^2+x-x-1\)
\(A=-1\)
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
Vì \(x=2017\Rightarrow x+1=2018\)
Thay \(x+1=2018\)vào biểu thức A ta được :
\(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+\left(x+1\right)\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1\)
\(=1\)
Đề:............
<=> - (1 - 2018x) + 2019x.(1 - 2018x) = 0
<=> (1 - 2018x).[(-1) + 2019x] = 0
Xét 2 trường hợp, ta có:
TH1: 1 - 2018x = 0 TH2: -1 + 2019x = 0
<=> 2018x = 1 <=> 2019x = 1
<=> x = 1/2018 <=> x = 1/2019
Vậy x = 1/2018; 1/2019
\(2018x-1+2019x\left(1-2018x\right)=0\)
\(-\left(1-2018x\right)+2019x\left(1-2018x\right)=0\)
\(\left(1-2018x\right)\left(-1+2019x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-2018x=0\\-1+2019x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2018}\\x=\frac{1}{2019}\end{cases}}}\)
THAY 2018 = xyz vào biểu thức
\(\frac{xyzx}{xy+xyzx+xyz}\) + \(\frac{y}{yz+y+xyz}\)+ \(\frac{z}{xz+z+1}\)
= \(\frac{xz}{1+xz+z}\)+ \(\frac{1}{z+1+xz}\)+ \(\frac{z}{xz+z+1}\)= \(\frac{xz+z+1}{xz+z+1}\)=\(1\)
Đặt \(A=\frac{2018x}{xy+2018x+2018}+\frac{y}{yzz+y+2018}+\frac{z}{xz+z+1}\)
Thay \(xyz=2018\)vào A ta được
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{1}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz+1+z}{xz+z+1}=1\)
x^4+2018x^2−2017x+2018
=(x^4+x)+(2018x^2−2018x+2018)
=x(x^3+1)+2018(x^2−x+1)
=x(x+1)(x^2−x+1)+2018(x^2−x+1)
=(x^2−x+1)[x(x+1)+2018]
=(x^2−x+1)(x^2+x+2018)
=(x^2−x+1)(x^2+x+2018)
thay xyz=2018 vào M ta có
\(M=\frac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+x+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+y\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+x+1}\)
\(=\frac{xz}{1+xz+y}+\frac{1}{z+1+xz}+\frac{z}{xz+1+xz}=\frac{xz+1+z}{z+1+xz}=1\)
Vậy M=1 với xyz=2018
Em chỉ làm đại thôi ạ, có gì sai mong chị bảo vì năm nay em mới lên lớp 7 :vv
\(M=\frac{2018x}{xy+2018x+2018}+\frac{y}{yz+y+2018}+\frac{z}{xz+z+1}\)
\(=\frac{2018x}{xy+2018x+2018}+\frac{xy}{xyz+xy+2018x}+\frac{xyz}{xyxz+xyz+xy}\)
\(=\frac{2018x}{xy+2018x+2018}+\frac{xy}{2018+xy+2018x}+\frac{2018}{xy+2018+2018x}\)
\(=\frac{2018x+xy+2018}{xy+2018x+2018}=1\)
Vậy M = 1.
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
@ 肖战Daytoy_1005 giup