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Bài 1 :
\(M+N\)
\(=\left(2xy^2-3x+12\right)+\left(-xy^2-3\right)\)
\(=2xy^2-3x+12-xy^2-3\)
\(=\left(2xy^2-xy^2\right)-3x+\left(12-3\right)\)
\(=xy^2-3x+9\)
\(h\left(x\right)+f\left(x\right)-g\left(x\right)=-2x^2-x+9\)
\(h\left(x\right)+\left(-5x^4+x^2-2x+6\right)-\left(-5x^4+x^3+3x^2-3\right)=-2x^2-x+9\)
\(h\left(x\right)-5x^4+x^2-2x+6+5x^4-x^3-3x^2-3=-2x^2-x+9\)
\(h\left(x\right)-\left(5x^4-5x^4\right)+\left(x^2-3x^2\right)-x^3-2x+\left(6-3\right)=-2x^2-x+9\)
\(h\left(x\right)-0-2x^2-x^3-2x+3=-2x^2-x+9\)
\(h\left(x\right)-x^3-2x^2-2x+3=-2x^2-x+9\)
\(h\left(x\right)+\left(-x^3-2x^2-2x+3\right)=-2x^2-x+9\)
\(h\left(x\right)=\left(-2x^2-x+9\right)-\left(-x^3-2x^2-2x+3\right)\)
\(h\left(x\right)=-2x^2-x+9+x^3+2x^2+2x-3\)
\(h\left(x\right)=\left(-2x^2+2x^2\right)-\left(x-2x\right)+\left(9-3\right)+x^3\)
\(h\left(x\right)=0+x+6+x^3\)
\(h\left(x\right)=x^3+x+6\)
d) Ta có : h(x) + f(x) - g(x) = -2x2 - x + 9
<=> h(x) = -2x2 - x + 9 - f(x) + g(x)
<=> h(x) = -2x2 - x + 9 - x2 + 2x + 5x4 - 6 + x3 - 5x4 + 3x2 - 3
<=> h(x) = x3 + x.
Vậy h(x) = x3 + x
Đăng từng bài thoy nha pn!!!
Bài 1:
Có : 2009 = 2008 + 1 = x + 1
Thay 2009 = x + 1 vào biểu thức trên,ta có :
x\(^5\)- 2009x\(^4\)+ 2009x\(^3\)- 2009x\(^2\)+ 2009x - 2010
= x\(^5\)- (x + 1)x\(^4\)+ (x + 1)x\(^3\)- (x +1)x\(^2\)+ (x + 1) x - (x + 1 + 1)
= x\(^5\)- x\(^5\)- x\(^4\)+ x\(^4\)- x\(^3\)+ x\(^3\)- x\(^2\)+ x\(^2\)+ x - x -1 - 1
= -2
1) Cho f(x) =0
=> x^2 + 6x +5 =0
x^2 +x +5x +5 = 0
x. ( x+1) + 5.(x+1) =0
(x+1) .(x+5) =0
=> x+1 =0 => x +5 =0
x =-1 x = -5
KL: x =-1 hoặc x =-5
bn lm như trên mk nha!!!!!
a)Xét các TH:
\(\cdot f\left(1\right)=1+1^2+1^3+...+1^{2020}\)(có 2020 số)
\(=1+1+...+1\)(có 2020 số 1)
\(=1\cdot2020=2020\)
\(f\left(x\right)=1+x+x^2+x^3+...+x^{2010}+x^{2011}\)
\(f\left(1\right)=1+1+1+1+....+1+1\)(2013 hạng tử)
\(f\left(1\right)=2013\)
\(f\left(-1\right)=1+\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+....+\left(-1\right)^{2010}+\left(-1\right)^{2011}\)
\(f\left(-1\right)=1+\left(-1\right)+1+\left(-1\right)+...+1+\left(-1\right)\)
\(f\left(-1\right)=\left[1+\left(-1\right)\right]+\left[1+\left(-1\right)\right]+....+\left[1+\left(-1\right)\right]+\left(-1\right)\)
\(f\left(-1\right)=-1\)
Nhầm :v làm lại
\(f\left(1\right)=1+1+1^2+1^3+....+1^{2010}+1^{2011}.\)(2012 số 1)
\(f\left(1\right)=1.2012=2012\)
\(f\left(-1\right)=1+\left(-1\right)+\left(-1\right)^2+....+\left(-1\right)^{2010}+\left(-1\right)^{2011}\)
\(f\left(-1\right)=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)+...+\left(1-1\right)\)(1006 cặp)
\(f\left(-1\right)=0\)
x=2020 nên x+1=2021
\(P\left(x\right)=x^{2021}-x^{2020}\left(x+1\right)+x^{2019}\left(x+1\right)-....+x\left(x+1\right)-2020\)
\(=x^{2021}-x^{2021}-x^{2020}+x^{2020}-...+x^2+x-2020\)
=x-2020=0
f(2020) = 20206 - 2021 × 20205 + 2021 × 20204 - 2021×20203 + 2021×20202 - 2021 × 2020 + 2021 = 1
Chúc bn học tốt !!!!!!!
Vì x = 2020
=> x + 1 = 2021
Khi đó f(2020) = x6 - (x + 1)x5 + (x + 1)x4 - (x + 1)x3 + (x + 1)x2 - (x + 1)x + (x + 1)
= x6 - x6 - x5 + x5 + x4 - x4 - x3 + x3 + x2 - x2 - x + x + 1
= 1