a: \(2^{333}=8^{111}< 9^{111}=3^{222}\)

a) Sử dụng máy tính cầm tay, ta có:

\(\left. \begin{array}{l}C_6^2 = 15\\C_6^4 = 15\end{array} \right\} \Rightarrow C_6^2 = C_6^4\)

b) Sử dụng máy tính cầm tay, ta có:

\(\left. \begin{array}{l}C_4^2 + C_4^3 = 6 + 4 = 10\\C_5^3 = 10\end{array} \right\} \Rightarrow C_4^2 + C_4^3 = C_5^3\)

a)

\(\begin{array}{l}{( - 3)^2}.{( - 3)^4} = 9.81 = 729\\ {( - 3)^6} = ( - 3).( - 3).( - 3).( - 3).( - 3).( - 3)\\ = 9.9.9 = 729\end{array}\)

Vậy \({( - 3)^2}.{( - 3)^4}\) = \({( - 3)^{6}}\)

b)

\(\begin{array}{l}0,6{}^3:0,{6^2} = 0,216:0,36 = 0,6\end{array}\)

Vậy \(0,6{}^3:0,{6^2}\) = \(0,{6}\)

b: \(\sqrt{3}-1=\sqrt{4-2\sqrt{3}}\)

mà \(4-3\sqrt{3}< 4-2\sqrt{3}\)

nên \(\sqrt{4-3\sqrt{3}}< \sqrt{3}-1\)

Đề này sai rồi bạn vì \(4-3\sqrt{3}< 0\)

em trả lời ccaua này hi vọng thầy còn nhớ em

a) -9/4<`1/3

c) 9/-5<7/-10

a) \(\dfrac{-9}{4}< 0\)

\(0< \dfrac{1}{3}\)

Do đó: \(\dfrac{-9}{4}< \dfrac{1}{3}\)

c) Đặt \(A=2^0+2^1+2^2+...+2^{50}\)

\(\Leftrightarrow2A=2^1+2^2+2^3...+2^{51}\)

\(\Leftrightarrow2A-A=2^1+2^2+2^3...+2^{51}\)\(-2^0-2^1-2^2-...-2^{50}\)

\(\Leftrightarrow A=2^{51}-2^0=2^{51}-1< 2^{51}\)

Vậy \(2^0+2^1+2^2+...+2^{50}< 2^{51}\)

a)Ta có: \(\hept{\begin{cases}2^{30}=\left(2^3\right)^{10}=8^{10}\\3^{30}=\left(3^3\right)^{10}=27^{10}\\4^{30}=\left(2^2\right)^{30}=2^{60}\end{cases}}\)và \(\hept{\begin{cases}3^{20}=\left(3^2\right)^{10}=9^{10}\\6^{20}=\left(6^2\right)^{10}=36^{10}\\8^{20}=\left(2^3\right)^{20}=2^{60}\end{cases}}\)

Mà \(8^{10}< 9^{10}\); \(27^{10}< 36^{10}\);\(2^{60}=2^{60}\)nên

\(8^{10}+27^{10}+2^{60}< 9^{10}+36^{10}+2^{60}\)

hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)

b) Ta có: \(4^{30}=2^{30}.2^{30}=8^{10}.4^{15}\)

\(3.24^{10}=3.8^{10}.3^{10}=3^{11}.8^{10}\)

Vì \(4^{15}>3^{11}\) nên \(8^{10}.4^{15}>3^{11}.8^{10}\)

hay \(2^{30}+3^{30}+4^{30}>3.24^{10}\)

em moi hoc lop 6

a >

b <

c >

d <

>

<

>

<

\(a:ta.c\text{ó}:BCNN:12\\ \dfrac{2}{3}=\dfrac{2\cdot4}{3\cdot4}=\dfrac{8}{12};\dfrac{1}{4}=\dfrac{1\cdot3}{4\cdot3}=\dfrac{3}{12}\\ v\text{ì }\dfrac{8}{12}< \dfrac{3}{12}n\text{ê}n\dfrac{2}{3}< \dfrac{1}{4}\\ b:ta.c\text{ó}:\\ 10=2\cdot5\\ 8=2^3\\ \Rightarrow BCNN=2^3\cdot5=8\cdot5=40\\ \dfrac{7}{10}=\dfrac{7\cdot4}{10\cdot4}=\dfrac{28}{40};\dfrac{7}{8}=\dfrac{7\cdot5}{8\cdot5}=\dfrac{35}{40}\\ v\text{ì }\dfrac{28}{40}< \dfrac{35}{40}n\text{ê}n\dfrac{7}{10}< \dfrac{7}{8}\\ c:ta.c\text{ó}:\\ 7=7;5=5\\ \Rightarrow BCNN=7\cdot5=35\\ \dfrac{6}{7}=\dfrac{6\cdot5}{7\cdot5}=\dfrac{30}{35};\dfrac{3}{5}=\dfrac{3\cdot7}{5\cdot7}=\dfrac{21}{35}\\ v\text{ì }\dfrac{30}{35}>\dfrac{21}{35}n\text{ê}n\dfrac{6}{7}>\dfrac{3}{5}\\ d:ta.c\text{ó}:\\ 21=3\cdot7\\ 72=2^3\cdot3^2\\ \Rightarrow BCNN=2^3\cdot3^2\cdot7=504\\ \dfrac{14}{21}=\dfrac{14\cdot24}{21\cdot24}=\dfrac{336}{504};\dfrac{60}{72}=\dfrac{60\cdot7}{72\cdot7}=\dfrac{420}{504}\\ v\text{ì }\dfrac{336}{504}< \dfrac{420}{504}n\text{ê}n\dfrac{14}{21}< \dfrac{60}{72}\)

Bài 1

a: 11/12=1-1/12

23/24=1-1/24

mà -1/12>-1/24

nên 11/12>23/24

b: -3/20=-9/60

-7/12=-35/60

mà -9>-35

nên -3/20>-7/12

A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60

Ta có: $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}+\frac{1}{2^{2022}}$

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}$

$\Rightarrow 2A-A=(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}})-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}+\frac{1}{2^{2022}})$

$\Rightarrow A=1-\frac{1}{2^{2022}}<1$

$\Rightarrow A<1\left(1\right)$

Lại có: $B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{17}{60}$

$\Rightarrow B=\frac{16}{15}$

$\Rightarrow B=1+\frac{1}{15}>1$

$\Rightarrow B>1\left(2\right)$

Từ $\left(1\right)$ và $\left(2\right)\Rightarrow A<B$

Vậy $A<B$