(2.x-16).2024=188.2024
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\(x+\left(x+1\right)+\left(x+2\right)+...+2023+2024=2024\)
\(\Rightarrow2023x+4090506=2024-2024-20232023\)
\(\Rightarrow x+4090506=-2023\)
\(\Rightarrow2023x=-2023-4090506\)
\(\Rightarrow2023x=-4092529\)
\(\Rightarrow x=-2023\).

\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)
\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)
Cộng vế:
\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)


a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110

a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
tìm giá trị lớn nhất của P = \(\dfrac{|x-2022|-|x-2023|+|x-2024|+2022}{|x-2022|+|x-2023|+|x-2024|}\)



\(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\)
\(\left|2x-1\right|=-\left(\dfrac{2}{3}-x\right)^{2024}\)
Vì \(VT\ge0;VP\le0\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}2x-1=0\\\dfrac{2}{3}-x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)(Loại)

Lời giải:
$y^2=36-8(x-2024)^2\leq 36$ (do $8(x-2024)^2\geq 0$)
$\Rightarrow y\leq 6$
Lại có: $y^2=36-8(x-2024)^2$ chẵn nên $y$ chẵn
$\Rightarrow y\in\left\{0; 2; 4; 6\right\}$
Nếu $y=0$ thì $8(x-2024)^2=36$
$\Rightarrow (x-2024)^2=\frac{36}{8}\not\in\mathbb{N}$ (loại)
Nếu $y=2$ thì $8(x-2024)^2=36-y^2=36-2^2=32$
$\Rightarrow (x-2024)^2=4\Rightarrow x-2024=\pm 2$
$\Rightarrow x=2026$ hoặc $x=2022$ (tm)
Nếu $y=4$ thì $8(x-2024)^2=36-4^2=20$
$\Rightarrow (x-2024)^2=\frac{20}{8}\not\in\mathbb{N}$ (loại)
Nếu $y=6$ thì $8(x-2024)^2=36-6^2=0$
$\Rightarrow x-2024=0$
$\Rightarrow x=2024$ (tm)
Vậy............

`@` `\text {Ans}`
`\downarrow`
`1.`
`13 \times 2525 - 25 \times 1313`
`= 13 \times 101 \times 25 - 25 \times 13 \times 101`
`= 13 \times 25 \times ( 101 - 101)`
`= 13 \times 25 \times 0`
`= 0`
`2.`
`(4 \times x - 2) \times (2024 - x) = x`
`4x(2024 - x) - 2(2024 - x) - x = 0`
`4x \times 2024 - 4x \times x - [ 2 \times 2024 + 2 \times (-x) ] = 0`
`8096 - 4x \times x - (4048 - 2x) = 0`
`8096 - 4x \times x - 4048 + 2x = 0`
`4048 - x(4x + 2) = 0`
`x(4x + 2) = 4048`
Bạn xem lại đề ;-; mình nghĩ lp 5 chưa hc mấy dạng ntnay ;-;.
(2x - 16) . 2024 = 188 . 2024
⇒ 2x - 16 = 188
2x = 188 + 16
2x = 204
x = 204 : 2
x = 102
(2x - 16) * 2024 = 188 * 2024
⇔ 2x - 16 = 188
⇔ 2x = 188 + 16
⇔ 2x = 204
⇔ x = 204 / 2
⇔ x = 102
Vậy x = 102.