Bài 2:

Ta có : \(2010=2011-1=x-1\)

Thay \(2010=x-1\) vào biểu thức A ,có :

\(x^{2011}-\left(x-1\right)x^{2010}-\left(x-1\right)x^{2009}-...-\left(x-1\right)x+1\)

\(=x^{2011}-x^{2011}+x^{2010}-x^{2010}+x^{2009}-...-x^2+x+1\)

\(=x+1\)

\(=2011+1=2012\)

Vậy giá trị biểu thức A là 2012

Bài 3:

\(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)

\(\Rightarrow a^2+2ab+b^2=c^2\)

\(\Rightarrow a^2+b^2-c^2=-2ab\left(1\right)\)

Tương tự :

\(a+b+c=0\)

\(\Rightarrow a+c=-b\)

\(\Rightarrow\left(a+c\right)^2=\left(-b\right)^2\)

\(\Rightarrow a^2+2ac+c^2=b^2\)

\(\Rightarrow a^2+c^2-b^2=-2ac\left(2\right)\)

\(a+b+c=0\)

\(\Rightarrow b+c=-a\)

\(\Rightarrow\left(b+c\right)^2=\left(-a\right)^2\)

\(\Rightarrow b^2+c^2-a^2=-2bc\left(3\right)\)

Từ (1)(2)(3)

\(\Rightarrow A=\dfrac{-ab}{2ab}+\dfrac{-bc}{2bc}+\dfrac{-ac}{2ac}\)

\(=\dfrac{-abc-abc-abc}{2abc}=\dfrac{-3abc}{2abc}=-\dfrac{3}{2}\)

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 :

\(Y=1+2+...+2^{2011}\)

\(2Y=2+2^2+...+2^{2012}\)

\(2Y-Y=\left(2+2^2+...+2^{2012}\right)-\left(1+2+...+2^{2011}\right)\)

\(Y=2^{2012}-1\)

\(\Rightarrow X=2^{2012}-2^{2012}+1\)

\(\Rightarrow X=1\)

\(\Rightarrow2010X=2010\)

x.x^4 nha

-Ta thấy \(x^4+x^2+1=x^4-x+x^2+x+1=\left(x^2-x\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2-x+1\right)\left(x^2+x+1\right)\)

Vậy PT sẽ thành

\(\frac{2010x\left(x^3+1\right)}{x\left(x^4+x^2+1\right)}+\frac{2010x\left(x^3-1\right)}{x\left(x^4+x^2+1\right)}=\frac{2011}{x\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow2.2010x^4=2011\Leftrightarrow x=...\)

gấp nha mấy bạn giups mình

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 #.

\(y=f\left(x\right)=\left|x^2-2011x+x-2011\right|\)

\(y=f\left(x\right)=\left|x\left(x-2011\right)+\left(x-2011\right)\right|\)

\(y=f\left(x\right)=\left|\left(x+1\right)\left(x-2011\right)\right|\)

Thay vào mà tính thôi bạn

ĐS: 2011x+1

Đúng ko ? :p

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