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\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+........+\frac{2}{x\left(x+1\right)}=\frac{2008}{2010}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+........+\frac{1}{x\left(x+1\right)}=\frac{2008}{4020}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{4020}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2008}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2008}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
=> x + 1 = 2010
=> x = 2010 - 1
=> x = 2009
\(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
e, \(A=\frac{2010^{2010}+1}{2010^{2011}}=>2010A=\frac{2010.\left(2010^{2010}+1\right)}{2010^{2011}}=\frac{2010^{2011}+2010}{2010^{2011}}=1+\frac{2010}{2010^{2011}}\)
\(B=\frac{2010^{2011}+1}{2010^{2012}}=>2010B=\frac{2010^{2012}+2010}{2010^{2012}}-1+\frac{2010}{2010^{2012}}\)
Vì \(2010^{2011}< 2010^{2012}=>\frac{2010}{2010^{2011}}>\frac{2010}{2010^{2012}}\)
=>2010A > 2010B
=> A > B
k cho mk nhé mấy câu trước dễ nên ko làm
\(\frac{13}{-9}>\frac{-24}{13}\)
\(-\frac{17}{82}>-\frac{26}{75}\)
\(\frac{-22}{35}< \frac{103}{-177}\)
\(\frac{2010^{2010}+1}{2010^{2011}}=\frac{2010^{2011}+1}{2010^{2012}}\)
xin lỗi bạn, bài cuối mình không chắc lắm!
à, bài đầu tiên mình lộn đề nhưng kết quả vẩn đúng nha!
sorry
mk nghĩ đây là toán 8.
\(Pt\Leftrightarrow\left(x-2010\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{72}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{8.9}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....-\frac{1}{9}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right).\frac{2}{9}=\frac{16}{9}\Leftrightarrow x-2010=8\Leftrightarrow x=2018.\text{ Vậy: x=2018}\)