Bài 6: So sánh phân số

\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)

Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

=>\(A>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)

a/b = 1/4

2:

\(GH=\dfrac{1}{2}EG=\dfrac{1}{2}\cdot10=5\left(cm\right)\)

Chu vi hình bình hành là:

\(2\cdot\left(10+5\right)=2\cdot15=30\left(cm\right)\)

1:

=>x/7=15/7

=>x=15

\(\dfrac{2x}{7}=\dfrac{75}{35}\)

→\(\dfrac{2x\times5}{7\times5}=\dfrac{75}{35}\)

→\(\dfrac{10\times x}{35}=\dfrac{75}{35}\)

→\(10\times x=75\)

→\(x=75:10=7,5=\dfrac{15}{7}\)

\(\dfrac{3}{-5}\) và \(\dfrac{-3}{4}\)

\(\dfrac{3}{-5}=\dfrac{3\times4}{-5\times4}=\dfrac{12}{-20};\dfrac{-3}{4}=\dfrac{-3\times \left(-5\right)}{4\times\left(-5\right)}=\dfrac{15}{-20}\)

Vì \(12< 15\) mà cả `2` đều là phân số âm nên \(\dfrac{12}{-20}>\dfrac{15}{-20}\)

`->` Vậy \(\dfrac{3}{-5}>\dfrac{-3}{4}\) 

3/5<3/4

=>-3/5>-3/4

\(\dfrac{3}{5}\)<\(\dfrac{3}{4}\)

--> \(\dfrac{3}{5}\)>-\(\dfrac{3}{4}\)

3/5<3/4

=>-3/5>-3/4

\(\dfrac{3}{-5}=\dfrac{3.4}{-5.4}=\dfrac{-12}{20}\)

\(\dfrac{3}{4}=\dfrac{3.5}{4.5}=\dfrac{15}{20}\)

Vì \(\dfrac{-12}{20}< \dfrac{15}{20}\) \(\Rightarrow\dfrac{3}{-5}< \dfrac{3}{4}\)

\(\dfrac{3}{-5}\) và \(\dfrac{-3}{4}\)

\(\dfrac{3}{-5}=\dfrac{-3}{5}\)

Ta thấy \(\dfrac{-3}{5}>\dfrac{-3}{4}\)

\(\Rightarrow\dfrac{3}{-5}>\dfrac{-3}{4}\)

a: 13/67=91/469>91/470

b: 141/813=22419/129267

159/904=22419/127464

=>141/813<159/904

a)

có `1/2=(1xx50)/(2xx50)=50/100`

vì `50<99` => `50/100<99/100`

`=> 1/2<99/100`

b)

ko so sánh được vì `1/0` không phải là phân số

a: 1/2=50/100<99/100

b: 1/0 ko xác định nha bạn

`a,` ta có :

`1/2 =(1xx50)/(2xx50)=50/100`

Vì `50 <99->1/2< 99/100`

`b,` Ta có :

`0/1=0;1/0=0`

`->0/1=1/0`

`@ Heavy hearted`