Có công thức: \(\frac{a}{b}< \frac{a+c}{b+c}\)

\(A=\frac{2016^{16}+1}{2016^{15}+1}< \frac{2016^{16}+1+2015}{2016^{15}+1+2015}=\frac{2016^{16}+2016}{2016^{15}+2016}=\frac{2016\left(2016^{15}+1\right)}{2016\left(2016^{14}+1\right)}=\frac{2016^{15}+1}{2016^{14}+1}=B\)

Vậy: \(A< B\)

A và B bằng nhau mà
 

=\(\frac{-2}{2017}\)

Tk mình nha!!!!

Kết quả bằng -2 /2017

\(\frac{5^{20}}{5^{16}+5^{15}.19}=\frac{5^{20}}{5^{15}\left(5+19\right)}=\frac{5^{20}}{5^{15}.24}=\frac{5^5}{24}\)

mình không biết làm

Ta có:

\(A=\frac{10^{15}+1}{10^{16}+1}\)

\(10A=\frac{10^{16}+10}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}\)

Ta so sánh \(10A\) và \(10B\)

Có: 

\(10A:\) Mẫu - tử = 9

\(10B:\) Mẫu - tử = 9

Lại có:

 \(\frac{10^{16}+10}{10^{16}+1}\) \(-1\)\(=\frac{9}{10^{16}+1}\)

\(\frac{10^{17}+10}{10^{17}+1}-1=\frac{9}{10^{17}+1}\)

Vì \(\frac{9}{10^{16}+1}\)\(>\frac{9}{10^{17}+1}\)nên \(10A>10B\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

Theo bải ra ta có:

A=\(\frac{10^{15}+1}{10^{16}+1}\)=> 10A =.\(\frac{10.\left(10^{15}+1\right)}{10^{16}+1}\)= \(\frac{10.10^{15}+1.10}{10^{16}+1}\)

                                      = \(\frac{10.10^{15}+10}{10^{16}+1}\)=\(\frac{10^{16}+1+9}{10^{16}+1}\)= \(1+\frac{9}{10^{16}+1}\)

B= \(\frac{10^{16}+1}{10^{17}+1}\)=> 10B = \(\frac{10.\left(10^{16}+1\right)}{10^{17}+1}\)=\(\frac{10.10^{16}+1.10}{10^{17}+1}\)

                                       = \(\frac{10.10^{16}+10}{10^{17}+1}\)= \(\frac{10^{17}+1+9}{10^{17}+1}\)= \(1+\frac{9}{10^{17}+1}\)

Vì 1=1 mà \(\frac{9}{10^{16}+1}\)>   \(\frac{9}{10^{17}+1}\)nên => 10A > 10B => A>B

Vậy A>B.

rõ ràng ta chỉ cần so sánh giữa \(15^{30}+16^{12}+17^{50}-16^8\) và \(17^{30}+16^8+15^{50}-16^{12}\)

Áp dụng tính chất nếu a>b thì a-b>0 ta được:

   \(15^{30}+16^{12}+17^{50}-16^8\)- \(\left(17^{30}+16^8+15^{50}-16^{12}\right)\)

= \(\left(17^{50}-17^{30}\right)+\left(16^{12}+16^{12}\right)+\left(15^{30}-15^{50}\right)-\left(16^8+16^8\right)\)

= \(\left(17^{50}-17^{30}\right)+\left(15^{30}-15^{50}\right)+2\left(16^{12}-16^8\right)\)

Vì 17^50 - 17^30 > l 15^30 - 15^50 l 

nên \(\left(17^{50}-17^{30}\right)+\left(15^{30}-15^{50}\right)>0\)

=>\(15^{30}+16^{12}+17^{50}-16^8\)> \(17^{30}+16^8+15^{50}-16^{12}\)

=> Phân số thứ nhất > 1 và p/s thứ hai < 1

Lúc này bạn tự so sánh nha

Đặt A = 1 + 2 + 4 + 8 + ... + 8192

2A = 2 + 4 + 8 + 16 + ... + 16384

2A - A = (2 + 4 + 8 + 16 + ... + 16384) - (1 + 2 + 4 + 8 + ... + 8192)

A = 16384 - 1

A = 16383

A=\(1+2^1+2^2+2^3+....+2^{13}.\)

2A=\(2+2^2+2^3+...+2^{14}.\)

2A-A=\(2^{14}-1\)

A=16383

\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^{19}}\)

\(\Leftrightarrow\)\(\frac{3^2.\left(2^2\right)^2.2^{16.2}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^{19}}\)

\(\Leftrightarrow\)\(\frac{2^4.2^{32}.3^2}{11.2^{35}-2^{76}}\)

\(\Leftrightarrow\)\(\frac{2^{36}.3^2}{2^{35}.\left(11-2^{41}\right)}\)

\(\Leftrightarrow\)\(\frac{2.3^2}{11-2^{41}}\)

Hết biết giải rồi 

\(\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}=\frac{3^2.2}{9}=\frac{3^2.2}{3^2}=2\)

Xét tử số (a)

27.4500+135.550.2

=(27.5).(4500:5)+135.(550.2)

=135.900+135.1100

=135.(900+1100)

=135.2000

=270000

Xét mẫu số(b)

2+4+6+8+...+16+18

Khoảng cách : 4-2=2

Số số hạng : (18-2):2+1=9(số)

Tổng:(18+2).9:2=90

\(\Rightarrow\frac{a}{b}=\frac{270000}{90}=30000\)

=> B = 30000

giúp mình với mai phải nộp rồi

\(A=4\left(\frac{1}{20}-\frac{1}{80}\right)=4.\frac{3}{80}=60\)

\(A=\frac{16}{20\cdot24}+\frac{16}{24\cdot28}+\frac{16}{28\cdot32}+...+\frac{16}{76\cdot80}\)

\(A=4\left[\frac{4}{20\cdot24}+\frac{4}{24\cdot28}+\frac{4}{28\cdot32}+...+\frac{4}{76\cdot80}\right]\)

\(A=4\left[\frac{1}{20}-\frac{1}{24}+...+\frac{1}{76}-\frac{1}{80}\right]\)

\(A=4\left[\frac{1}{20}-\frac{1}{80}\right]\)

\(A=4\left[\frac{4}{80}-\frac{1}{80}\right]=4\cdot\frac{3}{80}=\frac{4\cdot3}{80}=\frac{1\cdot3}{20}=\frac{3}{20}\)