\(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)

16 = 2⁴

(\(\dfrac{1}{16}\))²⁰⁰ = \(\dfrac{1}{2^{4.200}}\) = \(\dfrac{1}{2^{800}}\)= (\(\dfrac{1}{2}\))⁸⁰⁰

So sánh với (\(\dfrac{1}{2}\))¹⁰⁰⁰

Hai phân số cùng tử số, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn

Suy ra: (\(\dfrac{1}{16}\))²⁰⁰ > (\(\dfrac{1}{2}\))¹⁰⁰⁰

Ta có: \(\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2}\right)^{800}\)

mà \(\left(\dfrac{1}{2}\right)^{800}>\left(\dfrac{1}{2}\right)^{1000}\)

nên \(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)

a)

b)

+) Quy đồng mẫu số ba phân số $\frac{1}{4};\frac{3}{4};\frac{5}{8}$

$\frac{1}{4} = \frac{{1 \times 2}}{{4 \times 2}} = \frac{2}{8}$

$\frac{3}{4} = \frac{{3 \times 2}}{{4 \times 2}} = \frac{6}{8}$ ; Giữ nguyên phân số $\frac{5}{8}$

Vì $\frac{2}{8} < \frac{5}{8} < \frac{6}{8}$ nên $\frac{1}{4} < \frac{5}{8} < \frac{3}{4}$

Vậy các phân số xếp theo thứ tự từ bé đến lớn là: $\frac{1}{4};\,\,\frac{5}{8};\,\,\frac{3}{4}$

+) Quy đồng mẫu số ba phân số $\frac{2}{3};\,\,\frac{2}{9};\,\,\frac{5}{9}$

$\frac{2}{3} = \frac{{2 \times 3}}{{3 \times 3}} = \frac{6}{9}$ ; Giữ nguyên phân số $\frac{2}{9}$; $\frac{5}{9}$

Vì $\frac{2}{9} < \frac{5}{9} < \frac{6}{9}$ nên $\frac{2}{9} < \frac{5}{9} < \frac{2}{3}$

Vậy các phân số xếp theo thứ tự từ bé đến lớn là $\frac{2}{9};\,\,\frac{5}{9};\,\,\frac{2}{3}$

a) \(\dfrac{3}{4}=\dfrac{3\times4}{4\times4}=\dfrac{12}{16}\)

b) \(\dfrac{1}{3}=\dfrac{1\times3}{3\times3}=\dfrac{3}{9}\)

c) \(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18}\)

A<B

\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)

B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)

vậy A=B

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có

\(\dfrac{13^{16}+1}{13^{17}+1}< 1\Rightarrow\dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

vậy B<A

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có B<1 nên

\(\dfrac{13^{16}+1}{13^{17}+1}< \dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

Vậy B<A

\(7\dfrac{4}{5}và9\dfrac{1}{2}\\ Tacó:7< 9\\ \Rightarrow7\dfrac{4}{5}< 9\dfrac{1}{2}\\ 7\dfrac{1}{6}và3\dfrac{4}{5}\\ Tacó:7>3\\ \Rightarrow7\dfrac{1}{6}>3\dfrac{4}{5}\)

Câu cuối không phải hỗn số

C

1/4 = 2/8

C

A = \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\) + \(\dfrac{3}{25}\) +..............+ \(\dfrac{3}{(3n)^2}\)

A = ( \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\)+ \(\dfrac{3}{25}\)) +.....+ \(\dfrac{3}{(3n)^2}\)

A = 3. ( \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\))+............+ \(\dfrac{3}{(3n)^2}\)

A = 3.( \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + \(\dfrac{1}{5.5}\)) +............+ \(\dfrac{3}{(3n)^2}\)

Vì \(\dfrac{1}{2}\) > \(\dfrac{1}{3}\) > \(\dfrac{1}{4}\) > \(\dfrac{1}{5}\)Ta có : \(\dfrac{1}{2.2}>\dfrac{1}{2.3}>\dfrac{1}{3.3}>\dfrac{1}{3.4}>\dfrac{1}{4.4}>\dfrac{1}{4.5}>\dfrac{1}{5.5}>\dfrac{1}{5.6}\)

A > 3. ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)) + ............+ \(\dfrac{1}{(3n)^2}\)

A > 3. ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)) +.....+ \(\dfrac{1}{(3n)^2}\)

A > 3.( \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) +..............+ \(\dfrac{1}{(3n)^2}\)

A > 3. \(\dfrac{1}{3}\) +...............+ \(\dfrac{1}{(3n)^2}\)

A > 1 +..........+ \(\dfrac{1}{9n^2}\) > 1 

A > 1