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: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).

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

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

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

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

a) Ta có:

\( - \frac{1}{3} = \frac{{ - 5}}{{15}};\frac{{ - 2}}{5} = \frac{{ - 6}}{{15}}\)

Vì -5 > -6 nên \(\frac{{ - 5}}{{15}} > \frac{{ - 6}}{{15}}\) hay \( - \frac{1}{3}\) > \(\frac{{ - 2}}{5}\)

b) 0,125 < 0,13 vì chữ số hàng phần trăm của 0,125 là 2 nhỏ hơn chữ số hàng phần trăm của 0,13 là 3

c) Ta có:

\(\begin{array}{l} - 0,6 = \frac{{ - 6}}{{10}} = \frac{{ - 3}}{5} = \frac{{ - 9}}{{15}};\\\frac{{ - 2}}{3} = \frac{{ - 10}}{{15}}\end{array}\)

Vì -9 > -10 nên \(\frac{{ - 9}}{{15}} > \frac{{ - 10}}{{15}}\) hay - 0,6 > \(\frac{{ - 2}}{3}\)

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 \(0,6{}^3:0,{6^2}\) = \(0,{6}\)

a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)

 \({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)

Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);

b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)

\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)

Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)

c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)

\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)

Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).

d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)

Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).

(-2) ^4 . (-2) 65 và ( -2) ^ 12 : ( -2) ^3

=( -2) ^ 4+5 =(-2)^9 và (-2) ^12-3 = ( -2) ^9 

vậy ( -2) ^9 = (-2) ^9 

Nên (-2) ^4 .( -2) ^5 = ( -2) ^ 12 : ( -2) ^3

a) \(\left(-\dfrac{1}{3}\sqrt{63}\right)^2=\dfrac{1}{9}\cdot63=7\)

\(\left(-2\sqrt{2}\right)^2=8\)

mà 7<8

nên \(-\dfrac{1}{3}\sqrt{63}>-2\sqrt{2}\)

b) Ta có: \(\left(2\sqrt{55}\right)^2=4\cdot55=220\)

\(\left(\dfrac{3}{5}\sqrt{750}\right)=\dfrac{9}{25}\cdot750=270\)

mà 220<270

nên \(2\sqrt{55}< \dfrac{3}{5}\sqrt{750}\)

hay \(-2\sqrt{55}< -\dfrac{3}{5}\sqrt{750}\)