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A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
\(\text{đặt}k=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(K=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2017}\Rightarrow A=1\)
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)
\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
=\(\left(\frac{-1}{2}\right)\left(\frac{-2}{3}\right).....\left(\frac{-2014}{2015}\right)\left(\frac{-2015}{2016}\right)\)
=\(\frac{-1.-2.-3......-2014.-2015}{2.3.4......2015.2016}\)
=\(\frac{1}{2016}\)
Chúc bạn học tốt !
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{2015}-1\right)\left(\frac{1}{2016}-1\right)\)
=\(\frac{-1}{2}.\frac{-2}{3}......\frac{-2014}{2015}.\frac{-2015}{2016}\)
=\(\frac{1}{2016}\)(ta rút gọn tích)
k cho mình nha!
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
\(10^{2016}\)= 1000...00(mình ko cần biết cso bao nhiêu cx 0, nó là bài đánh lừa nhá bn)
\(2^3\)= 8
\(10^{2016}\) + 8= 10000...08
có 1+0+0+...+0+8=9. vậy số này chia hết cho 9
mà như bạn thấy số này là số dương nên số đó là số tự nhiên nhá