b: \(7\cdot2^{13}< 8\cdot2^{13}=2^{16}\)

d: \(3^{99}=\left(3^{33}\right)^3\)

\(11^{21}=\left(11^7\right)^3\)

mà \(3^{33}>11^7\)

nên \(3^{99}>11^{21}\)

Đặt \(A=\frac{15+\frac{15}{7}-\frac{15}{11}+\frac{15}{2009}-\frac{15}{13}}{\frac{4}{2009}-\frac{4}{13}+\frac{4}{7}-\frac{4}{11}+4}\)

\(=\frac{15\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}{4\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}\)

\(=\frac{15}{4}\)

Đặt \(B=\frac{5\cdot2010-1996}{14+4\cdot2010}\)

\(=\frac{5\left(1996+4\right)-1996}{14+4\cdot2010}\)

\(=\frac{5\cdot1996+20-1996}{14+4\left(1996+4\right)}\)

\(=\frac{4\cdot1996+20}{4\cdot1996+30}\)

\(\Rightarrow A\cdot B=\frac{4\cdot1996+20}{4\cdot1996+30}\cdot\frac{15}{4}=\frac{15\cdot4\left(1996+5\right)}{4\left(4\cdot1996+30\right)}=\frac{15\left(1996+5\right)}{4\cdot1996+30}=\frac{30015}{8004}\)

mặc dầu ko khoa học lắm nhưng mình thấy cũng được đấy

\(\frac{-15}{4}\times\frac{9}{11}+\frac{15}{4}\times\frac{15}{11}-\frac{15}{4}\times\frac{6}{11}\)

\(=\frac{15}{4}\times\left(\frac{-9}{11}+\frac{15}{11}-\frac{6}{11}\right)\)

\(=\frac{15}{4}\times0\)

\(=0\)

\(=\frac{15}{4}\times\left(-\frac{9}{11}\right)+\frac{15}{4}\times\frac{15}{11}+\frac{15}{4}\times\left(-\frac{6}{11}\right)\)

\(=\frac{15}{4}\times\left(-\frac{9}{11}+\frac{15}{11}+\left(-\frac{6}{11}\right)\right)\)

\(=\frac{15}{4}\times0\)

\(=0\)

Biết rồi mà đi đố người khác . Chịu chị luôn em lạy 2 tay

tìm n   N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên

giúp mik với sắp thi r

Bài 1 \(x=\frac{5}{26}\)

Bài 2 \(x=\frac{43}{20}\)

\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)

\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)

\(=\frac{2^9.3^9.-17}{1}\)

Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)

\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)

\(=\frac{3.2^{18}}{2^{35}.3^2}\)

\(=\frac{1}{2^{17}.3}\)

a,\(=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)

\(\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)

=-1+1+-2/11

=0+-2/11

=-2/11

b,\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{40}\right)+\frac{-5}{17}\)

=1+-1+-5/17

=0+-5/17

=-5/17

c,\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+-\frac{7}{9}\right)+\frac{16}{17}\)

=1+-1+16/17

=0+16/17

=16/17

d,\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

a.\(\frac{-5}{9}\)+\(\frac{8}{15}\)+\(\frac{-2}{11}\)+\(\frac{4}{-9}\)+\(\frac{7}{15}\)

=\(\frac{-5}{9}\)+\(\frac{4}{-9}\)+\(\frac{8}{15}\)+\(\frac{7}{15}\)+\(\frac{-2}{11}\)

=(\(\frac{-5}{9}\)+\(\frac{-4}{9}\))+(\(\frac{8}{15}\)+\(\frac{7}{15}\))+\(\frac{-2}{11}\)

=(-1)+1+\(\frac{-2}{11}\)

=0+\(\frac{-2}{11}\)

=\(\frac{-2}{11}\).