Bài 1:Cho đa thức f(x) = x^6 - 2020x^5 + 2020x^4 - 2020x^3 + 2020x^2 - 2020x + 2020
Tính f(2019)
Bài 2: Cho f(x) = -x^2 + bx + c có nghiệm là -2. Tính f(1) + f(-5)
Giúp mình nhanh nhé đang cần gấp
Cho f(x)= x^99-2020x^98+2020x^97+..............-2020x^2 + 2020-1 .Tính F(2019)?
f(x) = x^6 -2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020
Tính f(2019)
f(x) = \(\left(x^6-2019x^5\right)-\left(x^5-2019x^4\right)+\left(x^4-2019x^3\right)-\left(x^3-2019x^2\right)+\left(x^2-2019x\right)-\left(x-2019\right)+1\)
= \(x^5\left(x-2019\right)-x^4\left(x-2019\right)+x^3\left(x-2019\right)-x^2\left(x-2019\right)+x\left(x-2019\right)-\left(x-2019\right)+1\)
Thay x = 2019 vào f(x), ta có:
f(2019) = 0 + 0 + 0 + 0 + 0 +0 + 1 = 1
f(x)=x^99-2020x^98 + 2020x^97-2020x^96+........._2020x^2+2020x-1
Tính f(2019)
Bài làm:
Ta có: \(x=2019\Rightarrow2020=x+1\)
Thay vào ta được:
\(f\left(2019\right)=x^{99}-\left(x+1\right)x^{98}+\left(x+1\right)x^{97}-\left(x+1\right)x^{96}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(f\left(2019\right)=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}-x^{96}+...-x^3-x^2+x^2+x-1\)
\(f\left(2019\right)=x-1\)
Thay \(x=2019\)vào ta được:
\(f\left(2019\right)=2019-1=2018\)
Vậy f(2019) = 2018
\(f\left(x\right)=x^{99}-2020x^{98}+2020x^{97}-2020x^{96}+...-2020x^2+2020x-1\)
\(f\left(2019\right)=2019^{99}-2020.2019^{98}+2020.2019^{97}-...+2020.2019-1\)
Xét \(2020.2019^{98}=2019^{99}+2019^{98};2020.2019^{97}=2019^{98}+2019^{97}\)
\(2020.2019^{96}=2019^{97}+2019^{96};...;2020.2019=2019^2+2019\)
\(\Rightarrow f\left(2019\right)=2019^{99}-2019^{99}-2019^{98}+2019^{97}-2019^{97}-...+2019^2+2019-1\)
\(\Rightarrow f\left(2019\right)=2019-1=2018\). Vậy \(f\left(2019\right)=2018\)
Cho f(x)=x^100-2020x^99+2020x^98-...+2020x^2-2020x+2000
Tính f(2019)
\(f\left(2019\right)=x^{100}-\left(2019+1\right)x^{99}+\left(2019+1\right)x^{98}-....+\left(2019+1\right)x^2-\left(2019+1\right)x+2000\)
\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-...+\left(x+1\right)x^2-\left(x+1\right)x+2000\)
\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-...+x^3+x^2-x^2-x+2000\)
\(=-x+2000=-2019+2000\)
\(=-19\)
Tính giá trị các biểu thức sau:
D= \(4x^2-2x+3x\left(x-5\right)\)tại \(x=-1\)
E= \(x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x\) tại \(x=2019\)
F= \(x^{10}+20x^9+20x^8+...+20x^2+20x\) tại \(x=19\)
Mấy bạn giúp mk vs ai nhanh mk sẽ vote ạ các bạn làm đầy đủ cho mk nha mk cảm ơn nhìu :33
\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)
\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)
F(x)=x^2019-2020x^2018+2020x^2017-2020x^2016+...+2020x-2020 tại x= 2019
Cho x=2019
Tính A= x6-2020x5+2020x4-2020x3+2020x2-2020x+2020
2020.2019^5 = (2019+1).2019^5 = 2019^6+2019^5 làm tương tự với các x còn lại
A= 2019^6 - 2019^6 +.....-2019^2-2019 +2020 = 1 vậy A=1
ta có x = 2019 \(\Rightarrow\)2020 = x+1
thay 2020 = x+1 vào A ta có
\(A=x^6-\left(x+1\right).x^5+\left(x+1\right).x^4-...-\left(x+1\right).x+2020\)
\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2020\)
\(=-x+2020\)
\(=-2019+2020\)
\(=1\)
vậy A = 1
học tốt !!!
Bài 1: Tính bằng cách hợp lí:
a) \(\)\(2^1_{317}.\frac{3}{111}-3^{316}_{317}.\frac{1}{111}-\frac{4}{111.317}\)
b) \(x^9-2020x^8+2020x^7-2020x^6+2020x^5-2020x^4+2020x^3-2020x^2+2020x\) tại x=2019
GIÚP MIK VS!!!! MIK ĐAG CẦN GẤP.
b) Ta có : \(x=2019\) \(\Rightarrow x+1=2020\) Thay vào biểu thức ta được :
( Chỗ nào có 2020 thay thành x + 1 )
\(x^9-\left(x+1\right).x^8+\left(x+1\right).x^7-....-\left(x+1\right).x^2+\left(x+1\right).x\)
\(=x^9-x^9-x^8+x^8+x^7-...-x^3-x^2+x^2+x\)
\(=x\\ \)
\(=2019\)
Vậy : biểu thức trên bằng 2019 với x = 2019.
Rút gọn:
a) A=(5-2x)2-4x(x-5)
b) B= (4-3x)(4+3x)+(3x+1)2
c) C= (x+1)3-x(x2+3x+3)
d) D=(2021x-2020)2-2(2021x-2020)(2020x-2021)+(2020x-2021)
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)