CÂU 1 = -59/111

 CÂU 2 = 11/63

cảm ơn kết quả thì mik b òi nhưng mik cần cách làm

a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)

= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)

=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)

= \(\frac{-5}{12}-\frac{8}{5}\)

= \(\frac{\left(-25\right)-96}{60}\)

= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)

=\(\frac{121}{60}\)

b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)

= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)

=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)

= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)

=\(\frac{-29}{24}\)

c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{3}{5}\)

d)\(\frac{5^4.20^4}{25^5.4^5}\)

=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)

=\(\frac{100^4}{100^5}\)

=\(\frac{1}{100}\)

a)\(\frac{25.4-0,5\cdot40\cdot0,2\cdot20\cdot0,25}{1+2+8+...+129+156}\)

\(=\frac{100-100}{1+2+8+...+156}\)

\(=\frac{0}{1+2+8+...+156}\)

\(=0\)

b)\(\frac{0,5\cdot40-0,5\cdot20\cdot8\cdot0,1\cdot0,25\cdot10}{128:8.16.4\left(4+52:4\right)}=\frac{20-20}{128:8.16.4.\left(4+52:4\right)}=\frac{0}{128:8.16.4.\left(4+52:4\right)}=0\)

\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)

\(A=\frac{1.2}{3.4}\)

\(A=\frac{1}{6}\)

Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)

Vậy B > A

Theo đề bài, ta có:

\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)

\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)

\(A=\frac{1\times2}{3\times4}\)

\(A=\frac{1}{6}\)

Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)

Vậy \(B>A\)

dễ mak bn!!!

dễ thì làm hộ ik 

Ta có:

A = \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.5+15.20}\)(sửa 6,8 = 6 . 8 hay  6 x  8)

Ở phép tính này. Chúng ta thực hiện lượt các số giống nhau đi

Ta lại được:

\(\frac{1.2+2+3+4+5}{3+9.12+12+15.20}\)=  \(\frac{16}{423}\)

Rồi thực hiện so sánh. Tự làm nhá!

\(1+2+3+...+100=10x-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\left(6.20-3.40\right)\)

\(\Rightarrow5050=10x-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right).0\)

\(\Rightarrow5050=10x\)

\(\Rightarrow x=505\)

de thui

nhung ma bai nay dai qua

luc ranh mk lam cho

nha@@@@

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(\Rightarrow A=\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+...+\frac{1}{19}.\frac{1}{20}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)

\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\)                                

\(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)\(..........\)

Vay A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(A=\frac{1}{1}-\frac{1}{20}\)

\(A=\frac{19}{20}\)

sai de bai rui