\(\Leftrightarrow\left(\sqrt{x+2022}-\sqrt{y+2022}\right)+\left(x^3-y^3\right)=0\)

=>\(\dfrac{x-y}{\sqrt{x+2022}+\sqrt{y+2022}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

=>x-y=0

=>x=y

P=2x^2-5x^2+x^2+12x+2023

=-2x^2+12x+2023

=-2(x^2-6x-2023/2)

=-2(x^2-6x+9-2041/2)

=-2(x-3)^2+2041<=2041

Dấu = xảy ra khi x=3

Sửa đề: \(\frac{x-1}{2023}+\frac{x-2}{2022}+\cdots+\frac{x-2022}{2}=2022\)

Ta có: \(\frac{x-1}{2023}+\frac{x-2}{2022}+\cdots+\frac{x-2022}{2}=2022\)

=>\(\left(\frac{x-1}{2023}-1\right)+\left(\frac{x-2}{2022}-1\right)+\cdots+\left(\frac{x-2022}{2}-1\right)=2022-2022=0\)

=>\(\frac{x-2024}{2023}+\frac{x-2024}{2022}+\cdots+\frac{x-2024}{2}=0\)

=>\(\left(x-2024\right)\left(\frac{1}{2023}+\frac{1}{2022}+\cdots+\frac12\right)=0\)

=>x-2024=0

=>x=2024

A

chx chắc là A đâu, bạn cho mik bt dấu "=" xảy ra khi nào

24^5 K+1

@Bé Bin bạn có biết cách giải không zậy

|x - 2| + |y - 1| + (x - y - z)²⁰²² = 0 (1)

Do |x - 2| ≥ 0 với mọi x ∈ R

|y - 1| ≥ 0 với mọi x ∈ R

(x - y - z)²⁰²² ≥ 0 với mọi x ∈ R

(1) ⇒  |x - 2| = |y - 1| = (x - y - z)²⁰²² = 0

*) |x - 2| = 0

x - 2 = 0

x = 2

*) |y - 1| = 0

y - 1 = 0

y = 1

*) (x - y - z)²⁰²² = 0

x - y - z = 0

2 - 1 - z = 0

1 - z = 0

z = 1

⇒ C = 26x - 3y²⁰²² + z²⁰²³

= 26.2 - 3.1²⁰²² + 1²⁰²³

= 52 - 3 + 1

= 50

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