\(\frac{1}{50.48}-\frac{1}{48.46}-...-\frac{1}{4.2}\)
Các bn giúp mik bài tập này vs chiều nay mik phải nộp rồi . Các bn làm đc phần nào thì lm nhé .Thank nhiều
Bài 1 :Tính Nhanh
a) \(\frac{-5}{11}.\left(\frac{2222}{1010}+\frac{2222}{1515}+\frac{2222}{2121}+\frac{2222}{2828}+\frac{2222}{3636}+\frac{2222}{4545}\right)\)
b) \(\frac{1}{100}-\frac{1}{50.48}-\frac{1}{48.46}-...-\frac{1}{4.2}\)
c) \(\frac{\frac{1}{2}-\frac{1}{3}}{\frac{1}{3}-\frac{1}{4}}.\frac{\frac{1}{4}-\frac{1}{5}}{\frac{1}{5}-\frac{1}{6}}.....\frac{\frac{1}{98}-\frac{1}{99}}{\frac{1}{99}-\frac{1}{100}}\)
Rút gọn: \(A=\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^4+1}+\frac{4.3}{4.3^4+1}+...+\frac{4.k}{4.k^4+1}\)
\(\frac{4k}{4k^4+1}=\frac{4k}{4k^4+4k^2+1-4k^2}=\frac{4k}{\left(2k^2+1\right)^2-\left(2k\right)^2}=\frac{4k}{\left(2k^2+2k+1\right)\left(2k^2-2k+1\right)}=\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1}\)
\(=\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(=1-\frac{1}{2k\left(k+1\right)+1}=...\)
tìm số nguyên dương n thỏa mãn: \(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^4+1}+\frac{4.3}{4.3^4+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)
Ta có: \(4n^4+1=\left(4n^4+4n^2+1\right)-4n^2=\left(2n^2+2n+1\right)\left(2n^2-2n+1\right)\)
\(\frac{4n}{4n^4+1}=\frac{\left(2n^2+2n+1\right)-\left(2n^2-2n+1\right)}{\left(2n^2-2n+1\right)\left(2n^2+2n+1\right)}=\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\)
Thay vào ta có:
\(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^2+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow\frac{2n^2+2n}{2n^2+2n+1}=\frac{220}{221}\Rightarrow n=10\)
tinh
A= \(\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-...-\frac{1}{6.4}-\frac{1}{4.2}\)
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-....-\frac{1}{6.4}-\frac{1}{4.2}\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100.98}+\frac{1}{98.96}+....+\frac{1}{6.4}+\frac{1}{4.2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{98}+\frac{1}{98}-\frac{1}{96}+.....+\frac{1}{6}-\frac{1}{4}+\frac{1}{4}-\frac{1}{2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)\Rightarrow A=\frac{1}{100}-\frac{1}{100}+\frac{1}{2}\Rightarrow A=\frac{1}{2}\)
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-...-\frac{1}{6.4}-\frac{1}{4.2}\)
\(A=\frac{1}{100}-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{96.98}+\frac{1}{98.100}\right)\)
\(A=\frac{1}{100}-\frac{1}{2.2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\frac{49}{50}\)
\(A=\frac{2}{200}-\frac{49}{200}=-\frac{47}{200}\)
\(Tính:\)
\(A=\frac{1}{1.2}+\frac{1}{2.0}+\frac{1}{3.0}+\frac{1}{4.2}+\frac{1}{5.6}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{4}{5\cdot7}+.......+\frac{9}{27\cdot29}\)
thực hiện phép tính
a)\(\frac{27^4.2^3-3^{10}.4^3}{6^4.9^3}\)
b) \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
b) \(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+..+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}=-\frac{5.9.7.89}{5.4.7.7.89}=\frac{-9}{28}\)
TÍNH:
a)\(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}\)
b)\(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0\)
a,\(\frac{1}{100.98}\)-\(\frac{1}{98.96}\)-\(\frac{1}{96.94}\)-...-\(\frac{1}{4.2}\)
b,Cho A=\(2^{2016}-2^{2015}-2^{2014}-...-2-1.Tính\)A
Bài 1:
a, Rút gọn \(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
b, Tìm số tự nhiên thỏa mãn điều kiện:
\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\)
\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-1+\frac{1}{100}\)
\(A=\frac{2}{100}-1\)
\(A=\frac{1}{50}-1\)
\(A=\frac{-49}{50}\)
b,\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\) (1)
Đặt \(B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\)
\(\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)\)
\(=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}\)
\(2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)\)
\(=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)\)
\(B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2\)
\(=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3\)
Đặt \(C=2^3+2^4+2^5+2^n\)
\(\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)\)
\(C=2^4+2^5+2^6+...+2^{n+1}\)
\(2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)\)
\(C=2^{n+1}-2^3\)
Khi đó : \(B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3\)
\(=-2^{n+1}+2^3+n.2^{n+1}-2^3\)
=\(=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}\)
Vậy từ (1) ta có:\(\left(n-1\right),2^{n+1}=2^{n+34}\)
\(2^{n+34}-\left(n-1\right).2^{n+1}=0\)
\(2^{n+1}.[2^{33}-\left(n-1\right)]=0\)
Do đó \(2^{33}-n+1=0\)( Vì \(2^{n+1}\ne0\)với mọi \(n\))
\(n=2^{33}+1\)
Vậy \(n=2^{33}+1\)