a,  1 + 2 + 3 + 4 3 = 100;  1 3 + 2 3 + 3 3 + 4 3 = 100 nên  1 + 2 + 3 + 4 3 =  1 3 + 2 3 + 3 3 + 4 3

Vậy  1 + 2 + 3 + 4 3  =  1 3 + 2 3 + 3 3 + 4 3

b, 16.18.20.22 = (19 – 3)(19 – 1)(19 + 1)(19 + 3)

= (19 – 3)(19+3)(19 – 1)(19 + 1)

= ( 19 2  – 9)( 19 2  – 1)

=  19 4 - 9 . 19 2 - 19 2 + 9

=  19 4 - 10 . 19 2 + 9 <  19 4

Vậy 16.18.20.22 <  19 4

a)

Có: 

\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)

Vì \(\sqrt{117}>\sqrt{116}\)  nên \(3\sqrt{13}>2\sqrt{29}\)

b)

Có:

\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)

\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)

Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\)  nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)

c)

Có:

\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)

\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)

Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)

d)

Có:

\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)

\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)

lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)

 \(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)

a) x lớn hơn 120

b) x=8

2) a) 2002/2001 lớn hơn

b) 2015/2018 lớn hơn

c) 27/37 lớn hơn

Đúng thì like giúp mik nha. Thx bạn

bạn có chs tiktok ko

Ta có:

194<16.18 vì 16.18=288

=>194<16.18.20.22

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: \(\left(4+\sqrt{33}\right)^2=49+8\sqrt{33}=49+2\cdot\sqrt{528}\)

\(\left(\sqrt{29}+\sqrt{14}\right)^2=43+2\cdot\sqrt{29\cdot14}=43+2\cdot\sqrt{406}\)

mà 49>43 và 528>406

nên \(\left(4+\sqrt{33}\right)^2>\left(\sqrt{29}+\sqrt{14}\right)^2\)

=>\(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)