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P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2016}=\frac{2015}{2016}< \frac{1512}{2016}=\frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
mình gợi ý nè : bạn thử lấy T nhân với 2 xem ( cả hai vế nhé )
Nếu bạn không ra thì k cho mình đi mình trình bày cho đôn giản mà mỗi tội hơi dài một chút.
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
1/2T=2/22 +3/23 +4/24 +...+2017/22017
T-1/2T= (2/21+3/22+4/23+...+2017/22016)-(2/22+3/23+4/24+...+2017/22017)
1/2T=2/21+3/22+4/23+...+2017/22016-2/22-3/23-4/24-...-2017/22017
1/2T=1+(3/22-2/22)+(4/23-3/23)+...+(2017/22016-2016/22016)-2017/22017
1/2T=1+(1/22+1/23+1/24+...+1/22016)-2017/22017
xét A = 1/22+1/23+1/24+...+1/22016
phần này dễ bạn tự làm nhé
A=1/2-1/22016<1/2(vì 1/22016>0)
1/2T<1/21+1/2-(1/22016+2017/22017)
1/2T<3/2(vì 1/22016+2017/22017>0)
T<3/2:1/2
T<3
vậy T<3
1/2T=2/22 +3/23 +4/24 +...+2017/22017 T-1/2T= (2/21+3/22+4/23+...+2017/22016 )-(2/22+3/23+4/24+...+2017/22017 ) 1/2T=2/21+3/22+4/23+...+2017/22016 -2/22 -3/23-4/24 -...-2017/22017 1/2T=1+(3/22 -2/22 )+(4/23 -3/23 )+...+(2017/22016 -2016/22016 )-2017/22017 1/2T=1+(1/22+1/23+1/24+...+1/22016 )-2017/22017 xét A = 1/22+1/23+1/24+...+1/22016 phần này dễ bạn tự làm nhé A=1/2-1/22016<1/2(vì 1/22016>0) 1/2T<1/21+1/2-(1/22016+2017/22017 ) 1/2T<3/2(vì 1/22016+2017/22017>0) T<3/2:1/2 T<3
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right)..................\left(1^2-2016^6\right)}{2^2.3^2.4^2...........2016^2}\)
\(\Leftrightarrow A=\frac{\left(1-2\right)\left(1+2\right)\left(1-3\right)\left(1+3\right)........\left(1-2016\right)\left(1+2016\right)}{2^2.3^2..........2016^2}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(3\right)\left(-2\right)\left(4\right).............\left(-2015\right)\left(1017\right)}{\left(2.3.4......2016\right)\left(2.3.4.2016\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)......\left(-2015\right)\right]\left(3.4.....2017\right)}{\left(2.3.4....2016\right)\left(2.3.4...2017\right)}\)
\(\Leftrightarrow A=-\frac{1}{2016.2}=-\frac{1}{4032}>-\frac{2}{2016}\)
\(\Leftrightarrow A=-\frac{2}{2016}\)
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right)..........\left(1^2-2016^2\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(3\right)\left(-2\right)\left(4\right)....\left(-2015\right)\left(2017\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right).....\left(-2015\right)\right]\left(3.4.5...2017\right)}{\left(2.3.....2016\right)\left(2.3.4....2016\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)2017}{2016}=-\frac{2017}{2016}< \frac{1}{2}\)
=> A<1/2
a< 3/2