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 \(4-3\sqrt{3}< 0\)

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

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

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

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

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

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 \(3\sqrt{12}>2\sqrt{16}\)

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: \(\dfrac{-9}{4}< \dfrac{1}{3}\)

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

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

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

1.

a. -3a - 1 + 1 > -3b - 1 + 1 (cộng cả 2 vế cho 1)

  -3a . \(\left(\dfrac{-1}{3}\right)\) <  -3b . \(\left(\dfrac{-1}{3}\right)\) (nhân cả vế cho \(\dfrac{-1}{3}\) )

         a < b

b. 4a + 3 + (- 3) < 4b + 3 +(- 3) (cộng cả 2 vế cho -3)

   4a . \(\dfrac{1}{4}\) < 4b . \(\dfrac{1}{4}\) (nhân cả 2 vế cho \(\dfrac{1}{4}\) )

        a < b

2. 

a. Ta có: a < b 

3a < 3b ( nhân cả 2 vế cho 3)

3a - 7 < 3b - 7 (cộng cả 2 vế cho - 7 )

b. Ta có: a < b

-2a > -2b (nhân cả 2 vế cho -2)

5 - 2a > 5 - 2b ( cộng cẩ 2 vế cho 5)

c. Ta có: a < b 

2a < 2b (nhân cả vế cho 2)

2a + 3 < 2b + 3 (cộng cả 2 vế cho 3)

d. Ta có: a < b

3a < 3b (nhân cả 2 vế cho 3)

3a - 4 < 3b - 4 (cộng cả 2 vế cho -4)

Ta có: 3 < 4

đến đây ko bắt cầu qua đc chắc đề bài sai