\(2018x+4036=x^2+4x+4\)
Những câu hỏi liên quan
cho khai triển \(\left(2018x^2+x+2018\right)^{2018}=a_0+a_1x+a_2x^2+...+a_{4036}x^{4036}\)
tính \(T=a_0-a_2+a_4-...-a_{4032}+a_{4036}\)
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
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Tìm GTNN \(\dfrac{|x+x^2+x^3+...+x^{4036}|-|x+x^2+x^3+...+x^{4035}|+2018}{\sqrt{\sqrt{\sqrt{x}}}+\sqrt{\sqrt{x}}+\sqrt{x}+x+x^2+x^4+...+2^{4036}}\)
Phân tích đa thức thành nhân tử
a) 4x^16+81
b) x^4+2018x^2+2017x+2018
\(\text{a) }4x^{16}+81=4x^4+36x^2+81-36x^8\)
\(=\left(4x^{16}+36x^8+81\right)-36x^8\)
\(=\left[\left(2x^8\right)^2+2.2x^8.9+9^2\right]+\left(6x^4\right)^2\)
\(=\left(2x^8+9\right)^2-\left(6x^4\right)^2\)
\(=\left(2x^8+9-6x^4\right)\left(2x^8+9+6x^4\right)\)
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\(\text{b) }x^4+2018x^2+2017x+2018\)
\(=x^4+2018x^2+2018x-x+2018\)
\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)
\(=x\left(x^3-1\right)-2018\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2-x\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)
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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 \(x=2017\Rightarrow A=2017^4.0-2017^3.0+2017^2.0-2017.0+2017-1000\)
\(A=2017-1000=1017\)
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Cho x \(=\)2017, Tính giá trị biểu thức
A\(=\)\(x^9-2018x^8+2018x^7-2018x^6+2018x^5-2018x^4-2018x^3-2018x^2+2018x-2018\)
\(A=x^9-2018x^8+2018x^7-2018x^6+2016x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)
\(A=x^9-\left(2017+1\right)x^8+\left(2017+1\right)x^7-...+\left(2017+1\right)x-\left(2017+1\right)\)
\(A=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-...+\left(x+1\right)x-x-1\)
\(A=x^9-x^9-x^8+x^8+x^7-...+x^2+x-x-1\)
\(A=-1\)
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Mk sửa lại đề. bn tham khảo nha!!!
\(x=2017\)\(\Rightarrow\)\(x+1=2018\)
Ta có: \(A=x^9-2018x^8+2018x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)
\(=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\)
\(=x^9-x^9-x^8+x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-1\)
\(=1\)
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Xem thêm câu trả lời
Phân tích các đa thức sau thành nhân tử:
a, x^3 - x^2 - 8x +12
b, x^3 -4x^2 - 11x +30
c, 8x^2 +10x -3
d, 8x^2 -2x -1
e, x^3 +x -2
f, x^3 +3x^2 -4
g, x^3 y^3+x^2 y^2+4
h,x^3-2x-1
l,4x^4+y^4
k,x^5+x^4+1
m, 64x^4+y^4
n,81x^4+4
i, x^8+14x^4+1
p, a^3+b^3+c^3-3abc
q, x(x+4)(x+6)(x+10)+128
r, (2017x-1)^3-(2018x^3-2019)^3+(2018x^3-2017x-2018)^3
I : PTĐTTNT
a) \(\left(x^2-x-2\right)^2+\left(x-2\right)^2\)
b) \(x^4+2019x^2+2018x+2019\)
c) \(x^4+2x^3+5x^2+4x-5\)
help me
a) \(=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)
\(=x^4-2x^3-2x^2+8\)
\(=x^3\left(x-2\right)-2x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x^3-2x-4\right)\left(x-2\right)\)
\(=\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\right]\left(x-2\right)\)
\(=\left(x-2\right)^2\left(x^2+2x+2\right)\)
b) \(=x^4-x+2019\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)\
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c)\(x^4+2x^3+5x^2+4x-5\\=x^4+x^3+x^3-x^2+x^2+5x^2-x+5x-5\\ =x^2\left(x^2+x-1\right)+x\left(x^2+x-1\right)+5\left(x^2+x-1\right)=\left(x^2+x-1\right)\left(x^2+x+5\right)\)
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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
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