a: \(\left(1+2+3+4\right)^2=10^2=100\)

\(1^3+2^3+3^3+4^3=1+8+27+64=100\)

Do đó: \(\left(1+2+3+4\right)^2=1^3+2^3+3^3+4^3\)

b: \(19^4=130321\)

\(16\cdot18\cdot20\cdot22=126720\)

mà 130321>126720

nên \(19^4>16\cdot18\cdot20\cdot22\)

a) Ta có: (1 + 2 + 3 + 4)^2 = 10^2 = 100

1^3 + 2^3 + 4^3 = 1 + 8 + 64 = 9 + 64 = 75

Vì 100 > 75 nên (1 + 2 + 3 + 4)^2 > 1^3 + 2^3 + 4^3

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)