kết bạn nha 1+1=mấy
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P(x)=x(x+3)(x+1)(x+2)+1
P(x)=(x2+3x)(x2+3x+2)+1
Đặt x2+3x=a
Ta có:
P(x)=a(a+2)+1
P(x)=a2+2a+1
P(x)=(a+1)2
Vậy P(x)=(x2+3x)2


\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{2009}-\frac{1}{2009}\right)+\left(\frac{1}{2010}-\frac{1}{2010}\right)-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
~ Hok tốt ~
\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)
\(=\frac{1}{1}:\frac{2}{1}+\frac{1}{2}:\frac{3}{1}+\frac{1}{3}:\frac{4}{1}+...+\frac{1}{2009}:\frac{2010}{1}+\frac{1}{2010}:\frac{2011}{1}\)
\(=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{2009}\cdot\frac{1}{2010}+\frac{1}{2010}\cdot\frac{1}{2011}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2010}+\frac{1}{2010\cdot2011}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
Dấu " . " là dấu nhân nhé
1 + 1 = 2 kb nha
1+1=2
kb nha!!!