=>(x-2023)[(x-2023)^21-1]=0

=>x-2023=0 hoặc x-2023=1

=>x=2023 hoặc x=2024

2020/2021<1

2021/2022<1

2022/2023<1

2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023

=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4

\(5x^2+2y^2+6xy-8x-4y+4=0\)

\(\Leftrightarrow4x^2+x^2+y^2+y^2+2xy+4xy-8x-4y+4=0\)

\(\Leftrightarrow\left(4x^2+y^2+4+4xy-8x-4y\right)+\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left[\left(2x\right)^2+4xy+y^2-4\left(2x+y\right)+2^2\right]+\left(x+y\right)^2=0\)

\(\Leftrightarrow\left[\left(2x+y\right)^2-2\cdot\left(2x+y\right)\cdot2+2^2\right]+\left(x+y\right)^2=0\)

\(\Leftrightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)

Ta có: \(\left\{{}\begin{matrix}\left(2x+y-2\right)^2\ge0\forall x,y\\\left(x+y\right)^2\ge0\forall x,y\end{matrix}\right.\)  

\(\Rightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2\ge0\forall x,y\)

Mặt khác: \(\left(2x+y-2\right)^2+\left(x+y\right)^2=0\) 

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x+y-2=0\\x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot\left(-y\right)+y-2=0\\x=-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y+y-2=0\\x=-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=2\\x=-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\) 

Thay x,y vào P ta có:

\(P=2^{2023}+\left(-2\right)^{2023}=2^{2023}-2^{2023}=0\)

Vậy: ... 

...

olm sẽ hướng dẫn em làm bài này như sau:

Bước 1: em giải phương trình tìm; \(x\); y

Bước 2:  thay\(x;y\) vào P

(\(x-1\))²⁰²² + |y + 1| = 0

Vì (\(x-1\))²⁰²² ≥ 0 ∀ \(x\); |y + 1| ≥ 0  ∀ y

⇒ (\(x\) - 1)²⁰²²  + |y + 1| = 0

⇔ \(\left\{{}\begin{matrix}\left(x-1\right)^{2022}=0\\y+1=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\) (1) 

Thay (1) vào P ta có:

1²⁰²³.(-1)²⁰²² : )(2.1- 1)²⁰²² +  2023

=  1 + 2023

= 2024

a+b+c=12

mọi người hãy trình bầy rõ ra nhé 
em ko hiểu nên nếu nói tắt sẽ ko thể tiếp thu

\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)

\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)

\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)

\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)

Vì \(2024>2023=>2024^{2024}>2024^{2023}\)

\(=>2024^{2024}+1>2024^{2023}+1\)

\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)

\(=>A< B\)

\(#PaooNqoccc\)

Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$

$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$

$\Rightarrow 5a-a=5^{2024}-1$

$\Rightarrow 4a=5^{2024}-1$

$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)