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\(1^2-2^2+3^2-4^2+...+2009^2-2010^2+2011^2=\left(2011^2-2010^2\right)+\left(2009^2-2008^2\right)+...+\left(3^2-2^2\right)+1^2\)\(=\left(2011-2010\right)\left(2011+2010\right)+\left(2009-2008\right)\left(2009+2008\right)+...+\left(3-2\right)\left(3+2\right)+1\)
\(=1+2+3+4+...+2008+2009+2010+2011=\frac{2011\cdot2012}{2}\)
Lời giải:
$A=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+....+(2009^2-2010)^2+2011^2$
$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+....+(2009-2010)(2009+2010)+2011^2$
$=-(1+2+3+4+....+2009+2010)+2011^2$
$=-\frac{2010.2011}{2}+2011^2=2023066$
Ta có: \(A=1^2-2^2+3^2-4^2+...+2009^2-2010^2+2011^2\)
\(=-\left(1+2+3+4+...+2009+2010\right)+2011^2\)
\(=-2021055+4044121=2023066\)
Bài 4:
Ta có:
\(a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1\)
\(\Leftrightarrow\left(a^2-2b+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\)
Mà \(\hept{\begin{cases}\left(a-1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}}\)
Vậy \(\left(a,b,c\right)=\left(1;-2;\frac{1}{2}\right)\)
a, \(2x^2+2x+5x+5=2x\left(x+1\right)+5\left(x+1\right)=\left(2x+5\right)\left(x+1\right)\)
b,\(2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
c,\(x^3-3x^2+1-3x=\left(x^3+1\right)-3x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d,\(x^2-4x-5=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
e,\(\left(a^2+1\right)^2-4a^2=\left(a^2+1\right)^2-\left(2a\right)^2=\left(a^2-2a+1\right)\left(a^2+2a+1\right)=\left(a-1\right)^2\left(a+1\right)^2\)
\(\Leftrightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)=0\)
Tu \(a+b+c=1\Leftrightarrow a;b;c\le1\Leftrightarrow1-a;1-b;1-c\ge0\)
Tich tren >=0
Dau bang say ra khi:
\(a^2\left(1-a\right)=b^2\left(1-b\right)=c^2\left(1-c\right)=0\)
Ket hop voi a+b+c=1 ta thu dc a;b;c la hoan vi 0;0;1
\(P=1\)
12 - 22 + 32 - 42 + 52 - 62 + ... + 20092 - 20102 + 20112
= (1 - 2).(1 + 2) + (3 - 4).(3 + 4) + (5 - 6).(5 + 6) + ... + (2009 - 2010).(2009 + 2010) + 20112
= -3 + (-7) + (-11) + ... + (-4019) + 20112
= -(1 + 2 + 3 + 4 + 5 + 6 + ... + 2009 + 2010) + 20112
\(=-\frac{\left(1+2010\right).2010}{2}+2011^2\)
\(=-2011.1005+2011^2\)
= 2011.(2011 - 1005)
= 2011.1006
= 2023066