\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)

\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)

4-x=5 hoặc 4-x=-5

x=-1 hoặc x=9

a)  $\frac{3}{5} \times \frac{6}{7} = \frac{{3 \times 6}}{{5 \times 7}} = \frac{{18}}{{35}}$                                                 

 b) $\frac{2}{9} \times \frac{{18}}{3} = \frac{{2 \times 18}}{{9 \times 3}} = \frac{{36}}{{27}} = \frac{4}{3}$

c)  $\frac{5}{7} \times 12 = \frac{{5 \times 12}}{7} = \frac{{60}}{7}$                                            

d)  $21 \times \frac{2}{7} = \frac{{21 \times 2}}{7} = \frac{{42}}{7} = 6$

a,

\(\dfrac{89}{-13}< 0< \dfrac{1}{123}\\ \Rightarrow\dfrac{89}{-13}< \dfrac{1}{123}\)

Vậy \(\dfrac{89}{-13}< \dfrac{1}{123}\)

b,

\(\dfrac{-13}{15}>\dfrac{-15}{15}=-1=\dfrac{-30}{30}>\dfrac{-31}{30}\)

Vậy \(\dfrac{-13}{15}>\dfrac{-31}{30}\)

c,

\(\dfrac{125}{123}=\dfrac{123}{123}+\dfrac{2}{123}=1+\dfrac{2}{123}\\ \dfrac{99}{97}=\dfrac{97}{97}+\dfrac{2}{97}=1+\dfrac{2}{97}\)

Vì \(\dfrac{2}{97}>\dfrac{2}{123}\Rightarrow1+\dfrac{2}{97}>1+\dfrac{2}{123}\Leftrightarrow\dfrac{99}{97}>\dfrac{125}{123}\)

Vậy \(\dfrac{99}{97}>\dfrac{125}{123}\)

d,

\(\dfrac{125}{126}< \dfrac{126}{126}=1=\dfrac{986}{986}< \dfrac{987}{986}\)

Vậy \(\dfrac{125}{126}< \dfrac{987}{986}\)

\(\dfrac{-11}{-32}>\dfrac{16}{49}\)

\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)

D 

D

D 

Cô làm rồi em nhé:

https://olm.vn/cau-hoi/giup-em-voiii.8161766187032

Bài 1:

a: Sửa đề: 1/3^200

1/2^300=(1/8)^100

1/3^200=(1/9)^100

mà 1/8>1/9

nên 1/2^300>1/3^200

b: 1/5^199>1/5^200=1/25^100

1/3^300=1/27^100

mà 25^100<27^100

nên 1/5^199>1/3^300

a >

b <

c >

d <

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