\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

= \(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{\left(5-7.2\right).2^{28}.3^{18}}\)

= \(\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{-9.2^{28}.3^{18}}\)

= \(\frac{\left(5.2^2-2\right).2^{28}.3^{18}}{-9.2^{28}.3^{18}}\)

= \(\frac{18.2^{28}.3^{18}}{-9.2^{28}.3^{18}}\)

= 18/-9 = -2

\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

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

\(=\frac{2^{28}.3^{18}.\left(5.2^2-2\right)}{2^{28}.3^{18}.\left(5-7.2\right)}\)

\(=\frac{5.4-2}{5-14}\)\(=\frac{18}{-9}=-2\)

T**k mik nha!

\(A=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.6^{19}-7.2^{29}.3^{18}}\)

\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(A=\frac{2^{29}.3^{18}.\left(5.2.1-9\right)}{2^{28}.3^{18}.\left(5.1.3-7.2.1\right)}\)

\(A=\frac{2.1}{1}=2\)

thank you bạn nha, Le Quang Phúc. 

a)\(\frac{ }{ }\)

a) \(\frac{25}{1188}\)

b)\(\frac{4}{3}\)

c)\(\frac{17\times4}{-17}=\frac{4}{-1}=\frac{-4}{1}=-4\)

liệt kê các số dưới 11

bằng 2. nhấn đúng cho mih nha

tố uyên sai rồi

\(\frac{5.2^{30}.3^{18}-2^{^2}.3^{^{20}}.2^{^{27}}}{5.2^{^9}.6^{^{19}}-7.2^{^{29}}.27^{^6}}\)

\(=\frac{5.2^{^{30}}.3^{^{18}}-2^{^{29}}.3^{^{20}}}{5.2^{^9}.2^{^{19}}.3^{^{19}}-7.2^{^{29}}.\left(3^{^3}\right)^6}\)

\(=\frac{5.2.2^{^{29}}.3^{^{18}}-2^{^{29}}.3^{^2}.3^{^{18}}}{5.2^{^{28}}.3^{^{19}}-7.2^{^{29}}.3^{^{18}}}\)

\(=\frac{2^{^{29}}.3^{^{18}}\left(2.5-3^{^2}\right)}{5.2^{^{28}}.3.3^{^{18}}-7.2.2^{^{28}}.3^{^{18}}}\)

\(=\frac{2^{^{29}}.3^{^{18}}.\left(10-9\right)}{2^{^{28}}.3^{^{18}}.\left(5.3-7.2\right)}\)

\(=\frac{2^{^{29}}.3^{^{18}}}{2^{^{28}}.3^{^{18}}}=2\)

Học tốt

#Gấu

Ta có:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)

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

Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)

SAI RỒI ĐÁP ÁN LÀ N^2/(N+1)^2

b1 7/2³.3.5 để nguyên

9/2².3².10=1/2³.5(rút gọn 9 với 3² , 10 tách ra 2.5)=3/2³.3.5 (nhân cả tử vs mẫu với 3)

thế là được phân số 7/2³.3.5 và 3/2³.3.5

b2 rutgon phân so

-7.3+4.-6/-5.3+2.3=-7.3+(-8).3/-5.3+2.3=3(7+-8)/3(-5+3)=-1/2

ok tk nha hêhh