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

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 \(4-3\sqrt{3}< 0\)

a) \(\dfrac{4}{9}< \dfrac{4}{8}=\dfrac{1}{2}\)

b) \(\dfrac{5}{8}=\dfrac{15}{24}>\dfrac{14}{24}=\dfrac{7}{12}\)

a)
4/9 < 5/10 mà 5/10 =1/2
=> 4/9<1/2
 

a: \(\dfrac{4}{9}< \dfrac{1}{2}\)

b: \(\dfrac{5}{8}>\dfrac{7}{12}\)

c: \(-\dfrac{5}{8}>-\dfrac{17}{18}\)

d: \(-\dfrac{8}{15}>-\dfrac{2}{3}\)

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 >

b <

c >

d <

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

a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)

mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)

nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)

\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)

\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)

\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)

\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)

\(\left(\dfrac{1}{80}\right)^7\&\left(\dfrac{1}{243}\right)^6\)

\(\dfrac{1}{80}>\dfrac{1}{81}=\dfrac{1}{3^4}\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{3^4}\right)^7=\dfrac{1}{3^{7.4}}=\dfrac{1}{3^{28}}>\dfrac{1}{3^{30}}\)

\(=\dfrac{1}{\left(3^5\right)^6}=\left(\dfrac{1}{243}\right)^6\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)

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) Ta có: 12,(24) = 12,242424….

Đi từ trái sang phải, chữ số thập phân thứ 2 của 2 số khác nhau. Vì  6 > 4 nên 12,26 >12,(24)

b)

Đi từ trái sang phải, chữ số ở hàng chục của 2 số khác nhau. 3 > 2 nên 31,3(5) > 29,9(8)

a,Ta có:\(2=\sqrt{4}\)

Vì \(\sqrt{4}>\sqrt{3}\)

\(\Rightarrow2>\sqrt{3}\)

b,Ta có:\(6=\sqrt{36}\)

Vì \(\sqrt{36}< \sqrt{41}\)

\(\Rightarrow6< \sqrt{41}\)

c,Ta có:\(7=\sqrt{49}\)

Vì \(\sqrt{49}>\sqrt{47}\)

\(\Rightarrow7>\sqrt{47}\)

a) 2 =√4 > √3 ;    

b) 6=√36 < √41 ;    

c) 7=√49 > √47

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: \(\dfrac{-9}{4}< \dfrac{1}{3}\)