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\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)
\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)
\(=6:\left(-9\right)=-\frac{2}{3}\)
2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}-\frac{1}{3}\)
\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72\left(not27\right)}=\frac{5.18-10.27+15.36}{4\left(5.18-10.27+15.36\right)}=\frac{1}{4}\)
i) \(\frac{25^9}{5^{16}}-5^3:5\)
\(=\frac{\left(5^2\right)^9}{5^{16}}-5^2\)
\(=\frac{5^{18}}{5^{16}}-25\)
\(=5^2-25\)
\(=25-25\)
\(=0.\)
k) \(\frac{5}{7}-\left|\frac{2}{-7}\right|\)
\(=\frac{5}{7}-\left|\frac{-2}{7}\right|\)
\(=\frac{5}{7}-\frac{2}{7}\)
\(=\frac{3}{7}.\)
l) \(\frac{3^6.3^4}{9^3}\)
\(=\frac{3^{6+4}}{\left(3^2\right)^3}\)
\(=\frac{3^{10}}{3^6}\)
\(=3^4.\)
\(=81.\)
Chúc bạn học tốt!
3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b)\(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)\)
\(=-6\)
c)\(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)
\(=9.\left(-\frac{1}{3}\right)^3+9.\frac{1}{27}\)
\(=9.\left[\left(-\frac{1}{3}\right)^3+\frac{1}{27}\right]\)
\(=9.0=0\)
d)\(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14\)
\(B=\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{-6\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=-\frac{6}{9}=-\frac{2}{3}\)