cho da thucP(x)=1+x+x^2+x^3+...+x^2018 tinh P(1) P(-1) P(0) P(3)
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NT
0
TT
0
14 tháng 9 2019
a) (x+3)(x+5)=0
=>x+3=0 hoặc x+5=0
=>x=-3 hoặc -5
b) (x-1).5-1=0
=>5x-5-1=0
=>5x-6=0
=>5x=6
=>x=6/5
c)
HL
14 tháng 7 2016
a)\(f\left(x\right)=x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-\frac{1}{4}x+2x-3\)
\(=x^5-x^5+7x^4-9x^3-3x^2+2x^2+x^2-\frac{1}{4}x+2x-3\)
\(=7x^4-9x^3+\frac{7}{4}x-3\)
\(g\left(x\right)=5x^4-x^5+\frac{1}{2}x^2+x^5+x^2-4x^4-2x^3+3x^2+x^3-\frac{1}{4}\)
\(=-x^5+x^5+5x^4-4x^4-2x^3+x^3+\frac{1}{2}x^2+x^2+3x^2-\frac{1}{4}\)
\(=x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}\)
b)\(f\left(1\right)=7.1^4-9.1^3+\frac{7}{4}.1-3=7-9+\frac{7}{4}-3=-\frac{13}{4}\)
\(f\left(-1\right)=7.\left(-1\right)^4-9.\left(-1\right)^3+\frac{7}{4}.\left(-1\right)-3=7+9-\frac{7}{4}-3=\frac{45}{4}\)
\(g\left(1\right)=1^4-1^3+\frac{9}{2}.1^2-\frac{1}{4}=1-1+\frac{9}{2}-\frac{1}{4}=\frac{17}{4}\)
\(g\left(-1\right)=\left(-1\right)^4-\left(-1\right)^3+\frac{9}{2}.\left(-1\right)^2-\frac{1}{4}=1+1+\frac{9}{2}-\frac{1}{4}=\frac{25}{4}\)
HL
14 tháng 7 2016
c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)
\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)
f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)
\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)
T
1
KL
26 tháng 7 2017
A = (1 - 1/2) x (1 - 1/3) x (1 - 1/4) + ... + (1 - 1/2017) x (1 - 1/2018)
<=> A = 1/2 x 2/3 x 3/4 x ... x 2016/2017 x 2017/2018
<=> A = \(\dfrac{1\times2\times3\times...\times2016\times2017}{2\times3\times4\times...\times2017\times2018}\)
<=> A = \(\dfrac{1}{2018}\)
@Nguyễn Linh Ly
26 tháng 7 2017
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)\left(1-\dfrac{1}{2018}\right)\)\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2016}{2017}.\dfrac{2017}{2018}\)
\(A=\dfrac{1.2.3....2016.2017}{2.3.4....2017.2018}\)
\(A=\dfrac{1}{2018}\)
\(P\left(x\right)=1+x+x^2+...+x^{2018}\)
\(\Rightarrow P\left(1\right)=1+1+1+...+1=1.2019=2019\)
\(\Rightarrow P\left(-1\right)=1+\left(-1\right)+1+\left(-1\right)+...+1=1\)
\(\Rightarrow P\left(0\right)=1+0+0+...+0=1\)
\(\Rightarrow P\left(3\right)=1+3+3^2+...+3^{2018}\)
\(\Rightarrow3P\left(3\right)=3+3^2+...+3^{2019}\)
\(\Rightarrow2P\left(3\right)=3^{2019}-1\)
\(\Rightarrow P\left(3\right)=\frac{3^{2019}-1}{2}\)