Ta có : 

\(33^{52}=\left(33^4\right)^{13}=\left[\left(3.11\right)^4\right]^{13}=\left(3^4.11^4\right)^{13}=\left(11^3.891\right)^{13}\)

\(44^{39}=\left(44^3\right)^{13}=\left[\left(11.4\right)^3\right]^{13}=\left(11^3.4^3\right)^{13}=\left(11^3.64\right)^{13}\)

Do 891 > 64 => 33^52 > 44^39

33⁵² lớn hơn 44³⁹ 

\(a,\) Ta có : \(\hept{\begin{cases}2^{10}=2^{10}\\3^{12}=3^{10}.3^2\end{cases}}\)

Vì \(3^{10}>2^{10}\Rightarrow2^{10}< 3^{10}.3^2\)

Hay \(2^{10}< 3^{12}\)

\(b,\)  Ta có : \(\hept{\begin{cases}33^{52}=\left(33^4\right)^{13}=1185921^{13}\\44^{39}=\left(44^3\right)^{13}=85184^{13}\end{cases}}\)

Vì \(1185921^{13}>85184^{13}\)

Do đó : \(33^{52}>44^{39}\)

33⁵²<44³⁹

\(3^{-200}=\left(3^{-2}\right)^{100}=\left(\frac{1}{9}\right)^{100}\)

\(2^{-300}=\left(2^{-3}\right)^{100}=\left(\frac{1}{8}\right)^{100}\)

\(\frac{1}{9}< \frac{1}{8}\Rightarrow\left(\frac{1}{9}\right)^{100}< \left(\frac{1}{8}\right)^{100}\Rightarrow3^{-200}< 2^{-300}\)

\(33^{52}=\left(33^4\right)^{13}\)

\(44^{39}=\left(44^3\right)^{13}\)

\(33^4=\left(33^{\frac{4}{3}}\right)^3\approx106^3\)

\(106^3>44^3\Rightarrow\left(33^4\right)^{13}> \left(44^3\right)^{13}\Rightarrow33^{52}>44^{39}\)

giải 

a)3^-200<2^-300

b)33^52>44^39

Bài 1:

\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\) 

\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\) 

\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\) 

Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\) 

\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\) 

\(\dfrac{-18}{31}\) 

\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\) 

Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)

Bài 2:

a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\) 

b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)

5/7<7/6
17/13>45/52
4/3>99/100
34/56<42/53
97/252>85/163
73/97<91/95
 

5/7<7/6
17/13>45/52
4/3>99/100
34/56<42/53
97/252>85/163
73/97<91/95

ko có rảnh đâu nha con

b)39/52 và 98/112
98/112=98:14/112:14=7/8
39/52=39:13/52:13=3/4=3x2/4x2=6/8
vì 7<8=>39/52<98/112
đúng thì k cho mình nha

a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)