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\(\frac{2011.2012-1}{2001.2012}\)\(=\frac{2012.2013-1}{2012.2013}\)vì rút gọn hai phân số ta đều được kết quả là \(\frac{-1}{1}\)
Ta có \(\frac{2011.2012-1}{2011.2012}\)=\(\frac{2011^2+2011-1}{2011^2+2011}\)
Lại có \(\frac{2012.2013-1}{2012.2013}\)=\(\frac{2012^2+2012-1}{2012^2+2012}\)
Mặt khác có \(\frac{a}{b}\)<\(\frac{a+m}{b+m}\)(với a<b )
\(\Rightarrow\)\(\frac{2012^2+2012-1}{2012^2+2012}\)<...............
còn lại bn tự làm nha dễ lắm
a)N=\(\frac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}=\frac{2^{30}\cdot3^{18}\cdot5-2^{29}\cdot3^{20}}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}\)
\(=\frac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3\cdot3\right)}{2^{28}\cdot3^{18}}=\frac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}=2\)
Vậy N=2
\(A=\left(1-\frac{1}{2010}\right)-\left(1-\frac{1}{2011}\right)+\left(1-\frac{1}{2012}\right)-\left(1-\frac{1}{2013}\right)=-\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(A=-\frac{1}{2010.2011}-\frac{1}{2012.2013}\)
Vì 2010.2011 > 2009.2010 => \(\frac{1}{2010.2011}<\frac{1}{2009.2010}\)=> \(-\frac{1}{2010.2011}>-\frac{1}{2009.2010}\)
\(-\frac{1}{2012.2013}>-\frac{1}{2011.2012}\)
=> A > B
\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\)
\(=\frac{3}{8}\)
\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)
\(=1-0-0-0-...-0-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
\(d,\)giống câu a tự làm nha mỏi tay quá.
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)
=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)
=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)
Bài làm :
Ta có :
\(A=\frac{2008.2009+4018}{2010.2011-4020}\)
\(A=\frac{2008.2009+2009.2}{2010.2011-2010.2}\)
\(A=\frac{2009.\left(2008+2\right)}{2010.\left(2011-2\right)}\)
\(A=\frac{2009.2010}{2010.2009}=1\)
Vậy A=1 .
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(A=\frac{2008.2009+4018}{2010.2011-4020}\)
\(A=\frac{2008.2009+2009.2}{2010.2011-2010.2}\)
\(A=\frac{2009.\left(2008+2\right)}{2010.\left(2011-2\right)}\)
\(A=\frac{2009.2010}{2010.2009}=1\)
Vậy ....
a,Ta có \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}-\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)
\(=\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{2.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)}-\frac{3.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}{6.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}\)
=\(\frac{1}{2}-\frac{3}{6}=\frac{1}{2}-\frac{1}{2}=0\)
Vậy giá trị biểu thức bằng 0
b, Mình không hiểu cho lắm ạ , nếu ko phiền xin xem lại đầu bài ạ
\(\frac{2008\cdot2009+4018}{2010\cdot2011-4020}=\frac{2008\cdot2009+2009\cdot2}{2010\cdot2011-2010\cdot2}=\frac{\left(2008+2\right)\cdot2009}{2010\left(2011-2\right)}=\frac{2010\cdot2009}{2010\cdot2009}=1\)
b
2008.2009 + 4018 = 2008.2009 + 2.2009 0.25
= 2009.(2008+2) = 2009.2010 0.25
2010.2011-4020 = 2010.2011-2.2010 0.25
= 2010.(2011-2) = 2010.2009 0.25
⇒2008.2009 4018
2010.2011 4020
+
−
= 1