a: =17x100-120=1700-120=1580

a) \(17\cdot85+15\cdot17-120\)
\(=17\cdot\left(85+15\right)-120\)
\(=17\cdot100-120\)
\(=1700-120\)
\(=1580\)
b) \(27\cdot77+24\cdot27-27\)
\(=27\cdot\left(77+24-1\right)\)
\(=27\cdot100\)
\(=2700\)

c) \(2^3\cdot17-2^3\cdot14\)
\(=8\cdot17-8\cdot14\)
\(=8\cdot\left(17-14\right)\)
\(=8\cdot3\)
\(=24\)

\(\left(x-36\right):\left(2.3^2\right)=2^2.3\)

\(\Rightarrow\left(x-36\right):18=12\)

\(\Rightarrow\left(x-36\right)=12.18\)

\(\Rightarrow x-36=216\)

\(\Rightarrow x=216+36\)

\(\Rightarrow x=252\)

(x-36):(2.3²)=2.2³

⇒ (x-36):(2.9)=2.8

⇒ (x-36):18=16

⇒ x-36=18.16=288

⇒ x=288+36=324

\(\left(x-36\right):\left(2\cdot3^2\right)=2^2\cdot3\\ x-36=2^2\cdot3\cdot2\cdot3^2\\ x-36=3^3\cdot2^3\\ x-36=216\\ x=216+36\\ x=252\)

\(a,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\left(\frac{3}{7}\right)^2\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6=\left(\frac{9}{49}\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6\)

\(=\left(\left(\frac{9}{49}\right)^{10}:\left(\frac{9}{49}\right)^6\right).\frac{3}{7}=\left(\frac{9}{49}\right)^{10-6}.\frac{3}{7}=\left(\frac{9}{49}\right)^4.\frac{3}{7}=\left(\left(\frac{3}{7}\right)^2\right)^4.\frac{3}{7}\)

\(=\left(\frac{3}{2}\right)^8.\frac{3}{7}=\left(\frac{3}{2}\right)^9\)

\(b,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}=2+\left(\frac{1}{2}\right)^3=2+\frac{1}{6}=2\frac{1}{6}\)

\(c,\left(0,125\right)^3.51=\left(\frac{1}{8}\right)^3.51=\frac{1^3}{8^3}.51=\frac{1}{512}.51=\frac{51}{512}\)

a) 5¹⁴ = 5^2.7= (5²)⁷= 25⁷

3²¹= 3^3.7= (3³)⁷= 27⁷

Mà 25 < 27 

=> 5¹⁴<3²¹

b) 4³³= (4³)¹¹  = 64¹¹

3⁴⁴= (3⁴)¹¹ = 81¹¹

=> 4³³<3⁴⁴

c) 2¹⁶¹ > 2¹⁶⁰ = (2⁴)⁴⁰= 16⁴⁰

Mà 13⁴⁰< 16⁴⁰ < 2¹⁶¹

=> 13⁴⁰<2¹⁶¹

d) 17¹⁵= 17^3.5= (17³)⁵ = 4913⁵

31¹⁰= (31²)⁵= 961⁵

Mà 4913 > 961

=> 17¹⁵> 31¹⁰

e) 2²⁵⁵ = (2⁵)⁴⁵ = 32⁴⁵

3¹⁸⁰ = (3⁴)⁴⁵ = 81⁴⁵

=> 2²⁵⁵ < 3¹⁸⁰

g) 202³⁰³ = (202³)¹⁰¹ = (101³.2³)¹⁰¹= (101³.8)¹⁰¹= (101².8.101)¹⁰¹= (101².808)¹⁰¹

303²⁰² = (303²)¹⁰¹= (3².101²)¹⁰¹= (101².9) ¹⁰¹

Vì 101².9 < 101².808

=> 202³⁰³>303²⁰²

Ta có: \(A=2+2^3+2^5+...+2^{201}\) (Vì sai quy luật dãy nên mình đã sửa lại theo 1 đề khác, nếu cần bạn hãy ib với mình)

\(A=\left(2+2^3+2^5\right)+\left(2^7+2^9+2^{11}\right)+...+\left(2^{193}+2^{195}+2^{197}\right)+2^{199}+2^{201}\)

\(A=42+42\cdot2^6+...+42\cdot2^{192}+2^{199}+2^{201}\)

\(A=42\cdot\left(1+2^6+...+2^{192}\right)+2^{199}+2^{201}\)

Vì \(2^{199}+2^{201}\equiv2+2\equiv1\left(mod.3\right)\)

=> A chia 3 dư 1

Xin lỗi bị nhầm đề ạ

Vì \(2^{199}+2^{201}\equiv2+1\equiv3\left(mod.7\right)\)

=> A chia 7 dư 3

Bạn làm ơn ghi RÕ đề bài để mình giải nhé

Hì hì xin lỗi trả lời muộn thông cảm