Ta có: \(\widehat{BOC}=\widehat{AOD}\) (`2` góc đối đỉnh)

  \(=>\widehat{AOD}=45^o \)

Lại có: \(\widehat{BOC}+\widehat{DOB}=180^o \) (`2` góc kề bù)

  \(=>45^o +\widehat{DOB}=180^o  \)

 \(=>\widehat{DOB}=135^o \)

 Mà \(\widehat{DOB}=\widehat{AOC}\) (`2` góc đối đỉnh)

  \(=>\widehat{AOC}=135^o \)

\(1-\dfrac{a}{a}=1-1=0\)

\(\Rightarrow\) \(1-\dfrac{a}{a}=0\)

#Hphong

`|x+3|+|x-2|=4`

Ta có bảng:

`@` Với `x < -3` ta có:

    `-x-3+2-x=0`

`=>-2x=1`

\(=>x=\dfrac{-1}{2}\) (ko t/m)

`@` Với \(-3 \le x < 2\) ta có:

    `x+3+2-x=0`

`=>0x=-5` (Vô lí)

`@` Với \(x \ge 2\) ta có:

    `x+3+x-2=0`

`=>2x=-1`

\(=>x=\dfrac{-1}{2}\) (ko t/m)

Vậy ko có gtr của `x` t/m.

\(\dfrac{x+2}{2021}+\dfrac{x+3}{2020}=\dfrac{x+2021}{2}+\dfrac{x+2020}{3}\)

\(=>\dfrac{x+2}{2021}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2021}{2}+1+\dfrac{x+2020}{3}+1\)

\(=>\dfrac{x+2+2021}{2021}+\dfrac{x+3+2020}{2020}=\dfrac{x+2021+2}{2}+\dfrac{x+2020+3}{3}\)

\(=>\dfrac{x+2023}{2021}+\dfrac{x+2023}{2020}-\dfrac{x+2023}{2}-\dfrac{x+2023}{3}=0\)

\(=>(x+2023)(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2}-\dfrac{1}{3})=0\)

  Mà \(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2}-\dfrac{1}{3} \ne 0\)

  \(=>x+2023=0\)

 \(=>x=-2023\)

Thanks

\(2x^2-72=0\)

\(=>2x^2=72\)

\(=>x^2=72:2=36\)

\(=>x^2=6^2\) hoặc \(x^2=(-6)^2\)

\(=>x=6\)   hoặc \(x=-6\)

`2x^2 - 72=0`

`2x^2 = 0 +72`

`2x^2=72`

`x^2=72:2`

`x^2=36`

`x^2 = 6^2`

`=> x = 6`

`#LeMichael`

1/3+2/5x  (x+1)=1

= 2/5 x ( x + 1) = 1- 1/3

= 2/5 x ( x+ 1) = 2/3

= x+ 1 = 2/3 : 2/5

=x + 1 = 5/3

= x = 5/3 - 1

= x = 2/3

`1/3+2/5 . (x+1)=1`

`2/5 . (x+1)=1-1/3`

`2/5.  (x+1)=2/3`

`x+1=2/3 : 2/5`

`x+1=2/3xx5/2`

`x+1=5/3`

`x=5/3-1`

`x=5/3-3/3`

`x=2/3`

Vậy `x=2/3`

`#LeMichael`