a) 2,63=\(2\frac{63}{100}\)

b)\(\frac{2013}{100}\)=\(20\frac{13}{100}\)=20,13

c)\(\frac{-2}{5}=\frac{-14}{35}\)

\(\frac{-3}{7}=\frac{-15}{35}\)

vì\(\frac{-14}{35}>\frac{-15}{35}\Rightarrow\frac{-2}{5}>\frac{-3}{7}\)

THANK YOU

phần a mình thiếu \(2\frac{63}{100}=\frac{263}{100}\)nhé bạn

a. \(25.5^3.\frac{1}{625}.5^2=5^2.5^3.\frac{1}{5^4}.5^2=\frac{5^7}{5^4}=5^3\)

b. \(4.32:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{2^4}=\frac{2^4}{2^4}=1\)

c. \(5^2.3^5.\left(\frac{3}{5}\right)^2=5^2.3^5.3^2.\frac{1}{5^2}==\frac{5^2}{5^2}.3^7=3^7\)

d. \(\left(\frac{1}{7}\right)^2.\frac{1}{7}.49^2=\frac{1}{7^3}.7^4=\frac{7^4}{7^3}=7\)

a) $\frac{1}{3} + 1 = \frac{1}{3} + \frac{3}{3} = \frac{{1 + 3}}{3} = \frac{4}{3}$

b) $\frac{2}{5} + 2 = \frac{2}{5} + \frac{{10}}{5} = \frac{{2 + 10}}{5} = \frac{{12}}{5}$

c) $7 + \frac{1}{2} = \frac{{14}}{2} + \frac{1}{2} = \frac{{14 + 1}}{2} = \frac{{15}}{2}$

a, A = 3² . 4³ - 3² + 333

= 3² (4³ - 1) + 333

= 9 . 63 + 333

= 567 + 333

= 900 = 30²

b, B = 5 . 4³ + 2⁴ . 5 + 41

= 5 . 64 + 16 . 5 + 41

= 5 (64 + 16) + 41

= 400 + 41

= 441 = 21²

a) ta có: \(\frac{a}{4}=\frac{b}{5};\frac{b}{5}=\frac{c}{8}\)

\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{8}=\frac{5a}{20}=\frac{3b}{15}=\frac{3c}{24}\)

ADTCDTSBN

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bn tự áp dụng rùi tìm a;b;c nha

b) ta có: \(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}\)

ADTCDTSBN

có: \(\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}=\frac{3a+9-5b+10+7c-7}{15-15+49}\)

\(=\frac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{49}=\frac{86+12}{49}=\frac{98}{49}=2\)

=>...

c) ta cóL \(\frac{a}{7}=\frac{b}{6}\Rightarrow\frac{a}{35}=\frac{b}{30}\)

\(\frac{b}{5}=\frac{c}{8}\Rightarrow\frac{b}{30}=\frac{c}{48}\)

\(\Rightarrow\frac{a}{35}=\frac{b}{30}=\frac{c}{48}=\frac{2b}{60}\)

ADTCDTSBN

...

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