\(2020+2019+...+\left(x+2\right)+\left(x+1\right)+x=2020\)

\(\Leftrightarrow2019+2018+...+\left(x+1\right)+x=0\)

Xét dãy :\(A=2019+...+\left(x+1\right)+x\)

Dãy gồm \(\left(2020-x\right)\) số hạng

Có :\(A=\frac{\left(2019-x\right)\left(2020-x\right)}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2019+x=0\\2020-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2019\\x=2020\end{matrix}\right.\)

Có ai biết làm bài này không vậy? Làm ơn giúp mình với!

a ) 4 . ( x² + 1 ) = 0

            x² + 1   = 0 : 4

            x² + 1   = 0

                   x²  = 0 - 1

                   x²  = - 1

                   x²  = - 1² => x = - 1

Vậy x = - 1

Thế còn phần b

b ) - 2018 . ( X + 2019 ) = 0²⁰²⁰

    - 2018 . ( x + 2019 )  = 0

                   x + 2019   = 0 : ( - 2018 )

                   x + 2019   = 0

                              x   = 0 - 2019

                               x  = - 2019

Vậy x = - 2019

Lời giải:

$\frac{x+2}{2020}+\frac{x+2}{2020}=\frac{x+2019}{3}+\frac{x+2020}{2}$

$\frac{x+2}{2020}+1+\frac{x+2}{2020}+2=\frac{x+2019}{3}+1+\frac{x+2020}{2}+1$

$\frac{x+2022}{2020}+\frac{x+2022}{2020}=\frac{x+2022}{3}+\frac{x+2022}{2}$

$(x+2022)(\frac{1}{2020}+\frac{1}{2020}-\frac{1}{3}-\frac{1}{2})=0$

Dễ thấy $\frac{1}{2020}+\frac{1}{2020}-\frac{1}{3}-\frac{1}{2}<0$

Do đó: $x+2022=0$

$\Rightarrow x=-2022$

\(\Leftrightarrow\dfrac{x-2}{2020}-1+\dfrac{x-3}{2019}-1=\dfrac{x-2019}{3}-1+\dfrac{x-2020}{2}-1\)

=>x-2022=0

hay x=2022

a) Ta có:\(8\left(x-2019\right)^2⋮8\Rightarrow25-y^2⋮8\)\(\left(1\right)\)

Mặt khác: \(8\left(x-2019\right)^2\ge0\Rightarrow25-y^2\ge0\)\(\left(2\right)\)

Từ\(\left(1\right),\left(2\right)\)ta có: \(y^2=1;9;25\)

Xét:\(y^2=1\Rightarrow8\left(x-2019\right)^2=24\Rightarrow\left(x-2019\right)^2=3\left(ktm\right)\)

\(y^2=9\Rightarrow8\left(x-2019\right)^2=16\Rightarrow\left(x-2019\right)^2=2\left(ktm\right)\)

\(y^2=25\Rightarrow8\left(x-2019\right)^2=0\Rightarrow\left(x-2019\right)^2=0\Rightarrow x-2019=0\Rightarrow x=2019\left(tm\right)\)

Vậy \(y=5;x=2019\)

\(y=-5;x=2019\)