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

a, \(2^{332}>3^{223}\)

b,\(\frac{17^{17}+1}{17^{16}+1}=\frac{17^{18}+1}{17^{17}+1}\)

Bài 1:

Ta thấy A < 1

=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy A < B

Bài 2:

Ta thấy C < 1

=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)

Vậy C < D

Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(A=\frac{17^{18}-2}{17^{19}-2}< \frac{17^{18}-2-32}{17^{19}-2-32}=\frac{17^{18}-34}{17^{19}-34}=\frac{17\left(17^{17}-2\right)}{17\left(17^{18}-2\right)}=\frac{17^{17}-2}{17^{18}-2}=B\)

\(\Rightarrow\)\(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~

Công thức: \(\frac{a}{b}< \frac{a+c}{b+c}\left(\frac{a}{b}< 1;a;b;c\inℕ^∗\right)\)

Ta có:

\(A=\frac{17^{18}-2}{17^{19}-2}< B=\frac{17^{17}-2-32}{17^{18}-2-32}=\frac{17^{17}-34}{17^{18}-34}=\frac{17\left(17^{17}-2\right)}{17\left(17^{18}-2\right)}=\frac{17^{17}-2}{17^{18}-2}\)

Từ đó ta kết luận A < B

xem xong nhớ tích

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

\(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

vì \(\frac{16}{17^{18}+1}< \frac{16}{17^9+1}\)nên \(17B< 17A\)

\(=>B< A\)

áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có: \(A=\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}\)\(=\frac{17^{18}-17}{17^{20}-17}=\frac{17.\left(17^{17}-1\right)}{17.\left(17^{19}-1\right)}\)\(=\frac{17^{17}-1}{17^{19}-1}\)

\(\Rightarrow A< B\)

\(A=\frac{17^{18}-1}{17^{20}-1}\Rightarrow17^2A=\frac{17^{18}-1}{17^{18}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}\left(1\right)\)

\(B=\frac{17^{17}-1}{17^{19}-1}\Rightarrow17^2B=\frac{17^{17}-1}{17^{17}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(2\right)\)

\(\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}< \frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\Rightarrow1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}>1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(3\right)\)

Từ \(\left(1\right);\left(2\right)\&\left(3\right)\Rightarrow17^2A>17^2B\Leftrightarrow A>B.\)

\(A=\frac{115^{17}+2}{115^{17}-1}=\frac{115^{17}}{115^{17}-\left(2+1\right)}=\frac{115^{17}}{115^{17}-3}\)

\(\Rightarrow A=B\)

Ta có :

A = \(\frac{115^{17}+2}{115^{17}-1}\)= \(\frac{115^{17}-1+3}{115^{17}-1}\)= 1 + \(\frac{3}{115^{17}-1}\)

B = \(\frac{115^{17}}{115^{17}-3}\)= \(\frac{115^{17}-3+3}{115^{17}-3}\)= 1 + \(\frac{3}{115^{17}-3}\)

Vì 115¹⁷ - 1 > 115¹⁷ - 3 nên  \(\frac{3}{115^{17}-3}\) <  \(\frac{3}{115^{17}-1}\)

Vậy :1 + \(\frac{3}{115^{17}-1}\) < 1 + \(\frac{3}{115^{17}-3}\) hay A < B

Ta có:

\(A=\frac{17^{18}+1}{17^{19}+1}\)

\(17A=\frac{17\left(17^{18}+1\right)}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}\)

\(17A=\frac{(17^{19}+1)+16}{(17^{19}+1)}=1+\frac{16}{17^{19}+1}\)          (1)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

\(17B=\frac{17\left(17^{17}+1\right)}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}\)

\(17B=\frac{(17^{18}+1)+16}{(17^{18}+1)}=1+\frac{16}{17^{18}+1}\)          (2)

Từ (1) và (2) => \(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

=>\(17A< 17B\)

Hay \(A< B\)

Vậy \(A< B\)

Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy \(A< B\)

Chúc bạn học tốt ~