`M=1/2^2+1/3^2+1/4^2+...+1/2021^2`
Vì `1/2^2>1/(2.3)`
`1/(3^2)>1/(3.4)`
`....................`
`1/2021^2>1/(2021.2022)`
`=>M>1/(2.3)+1/(3.4)+............+1/(2021.2022)`
`=>M>1/2-1/3+1/3-1/4+..........+1/2021-1/2022`
`=>M>1/2-1/2022=505/1011=1/3+56/337>1/3(1)`
Vì `1/2^2<1/(1.2)`
`1/(3^2)<1/(2.3)`
`....................`
`1/2021^2<1/(2021.2020)`
`=>M<1/(1.2)+1/(2.3)+............+1/(2020.2021)`
`=>M<1-1/2+1/2-1/3+..........+1/2020-1/2021`
`=>M<1-1/2021<1(2)`
`(1)(2)=>1/3<M<1`

+Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3};\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4};\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5};...;\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}>\dfrac{1}{2021.2022}\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2021.2022}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2022}=\dfrac{1}{2}-\dfrac{1}{2022}=\dfrac{505}{1011}>\dfrac{1}{3}\left(1\right)\)+Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2021^2}< \dfrac{1}{2020.2021}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}< 1\left(2\right)\)Từ (1) và (2) suy ra: \(\dfrac{1}{3}< M< 1\)

Giải:

 \(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}\) 

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

...

\(\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}< \dfrac{1}{2020.2021}\) 

\(\Rightarrow M< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}\) 

\(\Rightarrow M< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\) 

\(\Rightarrow M< \dfrac{1}{1}-\dfrac{1}{2021}< 1\)

\(\Rightarrow M< 1\left(1\right)\) 

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\) 

...

\(\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}>\dfrac{1}{2021.2022}\) 

\(\Rightarrow M>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2021.2022}\) 

\(\Rightarrow M>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)  

\(\Rightarrow M>\dfrac{1}{2}-\dfrac{1}{2022}=\dfrac{505}{1011}=\dfrac{1}{3}+\dfrac{56}{337}>\dfrac{1}{3}\left(2\right)\)  

Vậy \(\dfrac{1}{3}< M< 1\) (đpcm)

Chúc bạn học tốt!

???

bn ơi???

Ta có :

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=1a^2+1b^2+1c^2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}\)

\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(=2^2=2=2+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

\(=\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{abc}=\dfrac{abc}{abc}\)

\(=a+b+c\)

\(=abc\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\\ \Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\\ \Rightarrow\dfrac{a+b+c}{abc}=1\\ \Rightarrow a+b+c=abc\left(dpcm\right)\)

Bài 2: 

a: A=(12,87+14,13)+(-14,7-37,3)

=27-52=-25

b: B=-1/3+2/5-2/3-3/5+1/5

=-1

`1`

`a, 1/2 +1/3= 3/6 + 2/6 =5/6`

`d, 1/3 +3/5= 5/15 + 9/15=14/15`

`c,4/5 +1/2= 8/10 + 5/10= 13/10`

`2`

`a,1/2 +1/4=2/4 +1/4=3/4`

`b, 2/3 +1/6 = 4/6+1/6=5/6`

`c, 7/12 +1/2=7/12+ 6/12= 13/12`

`3`

Giải

Cả `2` ngày đi tất cả số quãng đường là :

`1/4 +1/2 =1/4+ 2/4= 3/4 ( quãng đường)`

đ/s...

`@ yL`

b)\(98^2=\left(100-2\right)^2=10000-400+4=9604\)

Bài 2: 

a) Ta có: \(\left(x+3\right)^2-\left(x-3\right)^2-12x\)

\(=x^2+6x+9-x^2+6x-9-12x\)

=0

b) Ta có: \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)

\(=x^2-4x+4-x^2+4x-3\)

=-1