a: 6^9=(6^3)^3=216^3>15^3

B: 6^36=(6^2)^18=36^18>35^18

c: 7^18=(7^2)^9=49^9>30^9

d: 3^500=243^100

7^300=343^100

=>3^500<7^300

Ta có:\(\frac{3}{15}=\frac{3.2}{15.2}\)

\(\Rightarrow\frac{3}{15}=\frac{1}{5}\)

\(\Rightarrow\frac{6}{18}=\frac{1}{3}\)

\(\Rightarrow\frac{3}{15}< \frac{6}{18}\)

a) 3/15=6/30<6/18 suy ra 3/15<6/18

b)1/2=0.5

3/15=0.2

4/6=0.6

suy ra 4/5>1/2>3/15

a) ta có: \(\frac{3}{15}=\frac{1}{5};\frac{6}{18}=\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}< \frac{1}{3}\Rightarrow\frac{3}{15}< \frac{6}{18}\)

b) ta có: \(\frac{1}{2}=\frac{15}{30};\frac{3}{15}=\frac{1}{5}=\frac{6}{30};\frac{4}{6}=\frac{2}{3}=\frac{20}{30}\)

\(\Rightarrow\frac{6}{30}< \frac{15}{30}< \frac{20}{30}\)

\(\Rightarrow\frac{3}{15}< \frac{1}{2}< \frac{4}{6}\)

c) ta có: \(\frac{1345}{1612}=0,8;\frac{4524}{4625}=0,9\)

\(\Rightarrow0,8< 0,9\Rightarrow\frac{1345}{1612}< \frac{4524}{4625}\)

d) ta có: \(\frac{7}{5}>1;\frac{6}{9}=\frac{2}{3}< 1;\frac{4}{3}>1;\frac{7}{15}< 1;1=1\)

mà \(\frac{7}{5}=\frac{21}{15};\frac{4}{3}=\frac{20}{15}\)

\(\Rightarrow\frac{21}{15}>\frac{20}{15}\Rightarrow\frac{7}{5}>\frac{4}{3}\)

mà \(\frac{6}{9}=\frac{2}{3}=\frac{5}{15}\)

\(\Rightarrow\frac{10}{15}>\frac{7}{15}\Rightarrow\frac{6}{9}>\frac{7}{15}\)

\(\Rightarrow\frac{7}{5}>\frac{4}{3}>1>\frac{6}{9}>\frac{7}{15}\)

Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

a) Ta có:

\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)

\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)

Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)

b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)

Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)

- Ta có:

\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)

- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)

a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)

=>Bằng nhau

b: a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)

A>B

tk nha

\(a,\dfrac{5}{3}>\dfrac{3}{5};b,\dfrac{6}{11}< \dfrac{9}{5};c,\dfrac{6}{11}=\dfrac{6}{11};d,\dfrac{8}{9}< \dfrac{8}{5}\)

 \(a,>\)

\(b,< \)

\(c,=\)

\(d,< \)

>

<

=

<

1.

a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)

\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)

\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{6}{2}.\frac{10}{39}\)

\(=\frac{10}{13}\)

b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)

\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)

\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)

\(=\frac{3}{2}.\frac{5}{28}\)

\(=\frac{15}{56}\)

\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)

\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=3.\frac{10}{39}\)

\(=\frac{10}{13}\)

\(a.\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)

\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)

\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=3.\frac{10}{39}\)

\(=\frac{10}{13}\)

a) 36:(-6)<0

b) (-15):(-3)>(-63):7

a) \(\dfrac{6}{14}=\dfrac{6:2}{14:2}=\dfrac{3}{7}\)

\(\dfrac{3}{7}< \dfrac{4}{7}\)

b) \(\dfrac{6}{15}=\dfrac{6:3}{15:3}=\dfrac{2}{5}\)

\(\dfrac{3}{5}>\dfrac{2}{5}\)

c) \(\dfrac{10}{18}=\dfrac{10:2}{18:2}=\dfrac{5}{9}\)

\(\dfrac{5}{9}>\dfrac{2}{9}\)