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![](https://rs.olm.vn/images/avt/0.png?1311)
1)3x4-5x3y+6x2-10xy+2
=(3x4-5x3y)+(6x2-10xy)+2
=x3(3x-5y)+2x(3x-5y)+2
=x3.0+2x.0+2
=0+0+2
=2
2) x5-2010x4+2010x3-2010x2+2010x-2020
=x5-(2009+1)x4+(2009+1)x3-(2009+1)x2+(2009+1)x-2009-11
=x5-(x+1)x4+(x+1)x3-(x+1)x2+(x+1)x-x-11
=x5-x5-x4+x4+x3-x3-x2+x2+x-x-11
=-11
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
Hàm số: y=f(x)=| x2 - 2010x - 2011 |
* Với f(1) = | 12 - 2010 x 1 - 2011 |
= | 1 - 2010 - 2011 |
= | -4020 |
= 4020
Vậy với f(1) thì = 420
* Với f(-2010) = | ( -2010 )2 - 2010 x ( -2010 ) - 2011 |
= | -4040100 - ( -4040100 ) - 2011 |
= | 0 - 2011 |
= - 2011
Vậy với f(-2010) thì bằng -2011
# Chúc bạn học tốt #.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S=1+2+2^2+...+2^{100}\)
\(2S=2+2^2+2^3+...+2^{101}\)
\(2S-S=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)
\(S=2^{101}-1\)
b) \(X=2^{2012}-2^{2011}-...-2-1\)
\(X=2^{2012}-\left(1+2+...+2^{2011}\right)\)
Đặt \(X=2^{2012}-Y\)
Ta có :
\(Y=1+2+...+2^{2011}\)
\(2Y=2+2^2+...+2^{2012}\)
\(2Y-Y=\left(2+2^2+...+2^{2012}\right)-\left(1+2+...+2^{2011}\right)\)
\(Y=2^{2012}-1\)
\(\Rightarrow X=2^{2012}-2^{2012}+1\)
\(\Rightarrow X=1\)
\(\Rightarrow2010X=2010\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{-1}{2010}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\right)\)
\(=\dfrac{-1}{2010}-\left(1-\dfrac{1}{2010}\right)\)
\(=\dfrac{-1}{2010}-1+\dfrac{1}{2010}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2^{2011}+2^{2010}+...+2+1\)
\(\Leftrightarrow2A=2^{2012}+2^{2011}+...+2^2+2\)
\(\Leftrightarrow A=2^{2012}-1\)
\(x=2^{2012}-A=2^{2012}-2^{2012}+1=1\)
=>2010x=2010
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\frac{2010x+2680}{x^2+1}\)
\(=\frac{-335x^2-335+335x^2+2010x+3015}{x^2+1}\)
\(=-335+\frac{335\left(x+3\right)^2}{x^2+1}\ge-335\)
Dấu : \("="\)xảy ra khi và chỉ khi :
\(\frac{335\left(x+3\right)^2}{x^2+1}=0\)
\(\Leftrightarrow335\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)^2=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy GTNN của \(A\)là : \(-335\Leftrightarrow x=-3\)
Thay 2010 = x + 1 vào P ( x ),ta có :
\(^{x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x-1}\)
= x10 - x10 - x9 + x9 + x8 - x8 - x7 + ... + x3 + x2 - x2 + x - 1
= x + 1
= 2009 + 1
= 2010
Thay 2010 = x+ 1 vào P( x) ,có :
\(x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x-1\)
= \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2+x-1\)
= x+1
= 2009 + 1
= 2010