Bạn nào cho mk 1 ik, mk cho bn ý 3 ik luôn. Mk hứa nếu nói dối bạn có thể Báo cáo sai phạm mk.

45/27.............................49/35

quy đồng là ra ( hoặc tìm phân số trung gian)

k nhé

\(\frac{252525}{373737}=\frac{252525:10101}{373737:10101}=\frac{25}{37}\)

\(\frac{272727}{353535}=\frac{272727:10101}{353535:10101}=\frac{27}{35}>\frac{27}{37}>\frac{25}{37}\)

vậy \(\frac{252525}{373737}

\(\frac{252525}{373737}=\frac{25.10101}{37.10101}=\frac{25}{37}=\frac{875}{1295}\)

\(\frac{272727}{353535}=\frac{27.10101}{35.10101}=\frac{27}{35}=\frac{999}{1295}\)

Vì \(\frac{875}{1295}

\(\frac{252525}{373737}=\frac{252525:10101}{373737:10101}=\frac{25}{37};\frac{272727}{353535}=\frac{272727:10101}{353535:10101}=\frac{27}{35}\)

=>25/37<27/35

thì bn quy đòng mẫu số hai phân số ra

rùi xem p/s nào có tử số lớn hơn thì lớn hơn 

\(\frac{181818}{242424}=\frac{181818:10101}{242424:10101}=\frac{18}{24}=\frac{18:6}{24:6}=\frac{3}{4}\)

\(\frac{181818}{242424}=\frac{181818\div10101}{242424\div10101}=\frac{18}{24}\)

\(\frac{252525}{353535}=\frac{252525\div10101}{353535\div10101}=\frac{25}{35}\)

Ta có : \(\frac{18}{24}=\frac{18\times35}{24\times35}=\frac{630}{840};\frac{25}{35}=\frac{25\times24}{35\times24}=\frac{600}{840}\)

Vì \(\frac{630}{840}>\frac{600}{840}\)nên \(\frac{181818}{242424}>\frac{252525}{353535}\)

Vì

\(\frac{13}{27}\)= 0,481

\(\frac{16}{33}\)= 0,484

Nên :

\(\frac{13}{27}\)< \(\frac{16}{33}\)

a, Ta có: \(\frac{23}{27}>\frac{23}{29}>\frac{22}{29}\)

\(\Rightarrow\frac{23}{27}>\frac{22}{29}\)

b, Ta có:

\(\frac{12}{25}< \frac{1}{2}\)

\(\frac{25}{49}>\frac{1}{2}\)

\(\Rightarrow\frac{12}{25}< \frac{25}{49}\)

áp dụng công thức

a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)

\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)

Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)

b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)

\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)

Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)