a. 3^6 : 3^2 + 2^3 . 2^ 2 = 3^ (6-3) + 2^(3+2) = 3^3 + 2^5= 27 +  32 =59

b. ( 39. 42 - 37.42) : 42 = [ 42. ( 39-37) ] :42 = 39 -37 = 2

a) 59

b) 2

4 nhân 5 mũ 2 -3 nhân(24-9)

\(\dfrac{-2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)

\(=\dfrac{-2^6\cdot3^3\cdot5^3\cdot7}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)

\(=\dfrac{-2^6\cdot3^3\cdot5^3}{2^5\cdot3^2\cdot5^3}=-2\cdot3=-6\)

\(S=6+6^2+6^3+.......+6^{100}\)

\(=\left(6+6^2\right)+\left(6^3+6^4\right)+......+\left(6^{99}+6^{100}\right)\)

\(=6\left(6+6^2\right)+6^3\left(6+6^2\right)+.....+6^{99}\left(6+6^2\right)\)

\(=6.42+6^3.42+.........+6^{99}.42\)

\(=42\left(6+6^3+.........+6^{99}\right)⋮42\left(đpcm\right)\)

\(A=2+2^2+2^3+......+2^{60}\)

\(A=2^1+2^2+2^3+.......+2^{60}\)

\(A=\left(2^{60}-2^1\right):\left(2^2\right)\)

\(A=2^{58}\)

\(A=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\).

\(A=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=\left(2+2^2+2^3\right)+2^3\left(2+2^2+2^3\right)+...+2^{57}\left(2+2^2+2^3\right)\)

\(=14\left(1+2^3+...+2^{57}\right)⋮14\)

Ta thấy \(\left(3,14\right)=1\)nên \(A\)chia hết cho \(3.14=42\).

a)11⁶+11⁵=(..................1)+(..................1)=..........................2

Vì có chữ số tận cùng là 2 nên chia hết cho 4

Bài này thì chắc phải dùng đồng dư -_-

a) Ta có: 

11 đồng dư với -1 (mod 4) => 11⁵ đồng dư với (-1)⁵  = -1 (mod 4) => 11⁵ + 1 chia hết cho 4 

=> 11⁶ đồng dư với (-1)⁶ (mod 4)

=> 11⁶ đồng dư với 1 (mod 4)

=> 11⁶ - 1 chia hết cho 4

=> (11⁶ - 1) + (11⁵ + 1) chia hết cho 4

=> 11⁶ + 11⁵ chia hết cho 4

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(A=2.3+2^3.3+...+2^{59}.3\)

\(A=3\left(2+2^3+...+2^{59}\right)\)

 Vì \(3\left(2+2^3+...+2^{59}\right)⋮3\)

\(\Rightarrow A⋮3\)

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(A=2.7+2^4.7+...+2^{58}.7\)

\(A=7\left(2+2^4+...+2^{58}\right)\)

Vì \(7\left(2+2^4+...+2^{58}\right)⋮7\)

\(\Rightarrow A⋮7\)

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hỏi ngu thế

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

f) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Rightarrow x=2\)

g) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x=14\)

\(\Rightarrow x=\frac{7}{13}\)