Cho đa thức f(x)=x+x2 - x3 + x4 - ... + x2014 - x2015.
CMR:f(\(\dfrac{1}{5}\)) <\(\dfrac{1}{6}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
20 tháng 3 2021
Ta có: \(Q\left(x\right)=P\left(x\right)-H\left(x\right)\)
\(\Leftrightarrow H\left(x\right)=P\left(x\right)-Q\left(x\right)\)
\(\Leftrightarrow H\left(x\right)=1+x+2x^2+...+2015x^{2015}-x^{2015}-x^{2014}-...-x^2-x-1\)
\(\Leftrightarrow H\left(x\right)=2014x^{2015}+2013x^{2014}+2012x^{2013}+...+x^2\)
NK
0
CM
31 tháng 12 2017
Ta có: f(x) + h(x) = g(x)
Suy ra: h(x) = g(x) – f(x) = (x4 – x3 + x2 + 5) – (x4 – 3x2 + x – 1)
= x4 – x3 + x2 + 5 – x4 + 3x2 – x + 1
= ( x4 – x4) – x3 + (x2 + 3x2 ) – x + (5+ 1)
= -x3 + 4x2 – x + 6
CM
25 tháng 12 2017
Ta có: f(x) – h(x) = g(x)
Suy ra: h(x) = f(x) – g(x) = (x4 – 3x2 + x – 1) – (x4 – x3 + x2 + 5)
= x4 – 3x2 + x – 1 – x4 + x3 – x2 – 5
= (x4 – x4) + x3 – (3x2 + x2) + x - (1+ 5)
= x3 – 4x2 + x – 6
T
27 tháng 10 2023
a, Sửa đề:
\(3x^2-\sqrt3 x+\dfrac14(dkxd:x\geq0)\\=(x\sqrt3)^2-2\cdot x\sqrt3\cdot\dfrac12+\Bigg(\dfrac12\Bigg)^2\\=\Bigg(x\sqrt3-\dfrac12\Bigg)^2\)
b,
\(x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\)
c,
\(x^4+x^3+2x^2+x+1\\=(x^4+x^3+x^2)+(x^2+x+1)\\=x^2(x^2+x+1)+(x^2+x+1)\\=(x^2+x+1)(x^2+1)\)
d,
\(x^3+2x^2+x-16xy^2\\=x(x^2+2x+1-16y^2)\\=x[(x+1)^2-(4y)^2]\\=x(x+1-4y)(x+1+4y)\\Toru\)
15 tháng 8 2021
a: \(F\left(x\right)=x^5-3x^2+x^3-x^2-2x+5\)
\(=x^5+x^3-4x^2-2x+5\)
\(G\left(x\right)=x^5-x^4+x^2-3x+x^2+1\)
\(=x^5-x^4+2x^2-3x+1\)
b: Ta có: \(H\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(=x^5+x^3-4x^2-2x+5+x^5-x^4+2x^2-3x+1\)
\(=2x^5-x^4+x^3-2x^2-5x+6\)
\(f\left(x\right)=x+x^2-x^3+x^4-...+x^{2014}-x^{2015}\)
\(f\left(\dfrac{1}{5}\right)=\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-...+\dfrac{1}{5^{2014}}-\dfrac{1}{5^{2015}}\)
\(5f\left(\dfrac{1}{5}\right)=1+\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-...+\dfrac{1}{5^{2013}}-\dfrac{1}{5^{2014}}\)
\(5f\left(\dfrac{1}{5}\right)+f\left(\dfrac{1}{5}\right)=\left(1+\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-...+\dfrac{1}{5^{2013}}-\dfrac{1}{5^{2014}}\right)+\left(\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-...+\dfrac{1}{5^{2014}}-\dfrac{1}{5^{2015}}\right)\)
\(6f\left(\dfrac{1}{5}\right)=1-\dfrac{1}{5^{2015}}\Leftrightarrow f\left(\dfrac{1}{5}\right)=\dfrac{1}{6}-\dfrac{1}{6.5^{2015}}< \dfrac{1}{6}\left(đpcm\right)\)