Cách 1:

\(2^2:2^4=2^{2-4}=2^{-2}=\dfrac{1}{4}\)

Cách 2:

\(2^2:2^4=4:16=\dfrac{4}{16}=\dfrac{1}{4}\)

\(2016:\left[25-\left(3x+2\right)\right]=32\cdot7\\ \Rightarrow2016:\left[25-\left(3x+2\right)\right]=224\\ \Rightarrow25-3x-2=2016:224\\ \Rightarrow23-3x=9\\ \Rightarrow3x=23-9\\ \Rightarrow3x=14\\ \Rightarrow x=\dfrac{14}{3}\)

\(\dfrac{1}{4}\) gấp \(\dfrac{1}{8}\) số lần là:

\(\dfrac{1}{8}:\dfrac{1}{4}=\dfrac{1}{8}\times4=\dfrac{1}{2}\)

Vậy: \(\dfrac{1}{4}\) gấp \(\dfrac{1}{8}\) là \(\dfrac{1}{2}\) lần

Số tự nhiên liền sau thì ta \(+1\)

\(17+1\rightarrow18\)

\(99+1\rightarrow100\)

\(a+1\rightarrow a+1\)

Số tự nhiên liền trước thì ta \(-1\)

\(35-1\rightarrow34\)

\(1000-1\rightarrow999\)

\(b-1\rightarrow b-1\)

\(a,\left(9x^2y^3+6x^3y^2-4xy^2\right):3xy^2\\ =9x^2y^3:3xy^2+6x^3y^2:3xy^2-4xy^2:3xy^2\\ =3xy+2x^2-\dfrac{4}{3}\\ b,\dfrac{1}{2}xy\left(x^5-y^3\right)-x^2y\left(\dfrac{1}{4}x^4-y^3\right)\\ =\dfrac{1}{2}xy\cdot x^5-\dfrac{1}{2}xy\cdot y^3-x^2y\cdot\dfrac{1}{4}x^4+x^2y\cdot y^3\\ =\dfrac{1}{2}x^6y-\dfrac{1}{2}xy^4-\dfrac{1}{2}xy^4-\dfrac{1}{4}x^6y+x^2y^4\\ =\dfrac{1}{4}x^6y-\dfrac{1}{2}xy^4+x^2y^4\)

\(\left(5-x\right)^2+\left(x+5\right)^2-2\cdot\left(x+5\right)\cdot\left(x-5\right)\\=\left(x-5\right)^2-2\cdot\left(x+5\right)\cdot\left(x-5\right)+\left(x+5\right)^2\\ =\left(x-5-x-5\right)^2\\ =\left(-10\right)^2\\ =100\)

\(a,2\cdot2\cdot2\cdot3\cdot3\cdot3\cdot3=2^3\cdot3^4\\ b,5\cdot5\cdot5\cdot5\cdot4\cdot4\cdot4=5^4\cdot4^3\\ c,7\cdot7\cdot7\cdot7\cdot6\cdot6\cdot6\cdot6=7^4\cdot6^4\\ d,8\cdot8\cdot6\cdot6\cdot6\cdot7\cdot7\cdot7=8^2\cdot6^3\cdot7^3\)

\(A=n^3+\left(n+1\right)^3+\left(n+2\right)^3\\ =n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8\\ =\left(n^3+n^3+n^3\right)+\left(3n^2+6n^2\right)+\left(3n+12n\right)+\left(1+8\right)\\ =3n^3+9n^2+15n+9\\ =3\left(n^3+3n^2+5n+3\right)⋮3\cdot3\left(đpcm\right)\)

\(921\times\dfrac{8}{11}=\dfrac{921\times8}{11}=\dfrac{7368}{11}\)

\(27^6:3=\left(3^3\right)^6:3=3^{18}:3=3^{18-1}=3^{17}\)