vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)

mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)

=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

đợi tý

Đã trả lời rồi còn độ tí đồ ngull

\(\dfrac{x+1}{x^2+2022}\) là số nguyên thì:

\(\left(x+1\right)⋮\left(x^2+2022\right)\\ \Rightarrow\left[\left(x+1\right)\left(x-1\right)\right]⋮\left(x^2+2022\right)\\ \Rightarrow\left(x^2+x-x-1\right)⋮\left(x^2+2022\right)\\ \Rightarrow\left(x^2-1\right)⋮\left(x^2+2022\right)\\ \Rightarrow\left(x^2+2022-2023\right)⋮\left(x^2+2022\right)\)

 \(Mà.\left(x^2+2022\right)⋮\left(x^2+2022\right)\\ \Rightarrow2023⋮\left(x^2+2022\right)\\ \Rightarrow x^2+2022\inƯ\left(2023\right)\\ \Rightarrow x^2+2022\in\left\{-289;-119;-17;-7;-1;-2023;1;7;17;119;289;2023\right\}\)

Ta có: \(x^2+2022\ge0\Rightarrow x^2+2022=2023\Rightarrow x^2=1\Rightarrow x=\pm1\)

Vậy \(x=\pm1\) thì biểu thức trên là số nguyên

=)) của shop t edit nên ko cs link 

Sửa: \(Đk:x\ge0\)

\(C=1-\dfrac{1}{\sqrt{x}+2022}\ge1-\dfrac{1}{0+2022}=\dfrac{2021}{2022}\\ C_{min}=\dfrac{2021}{2022}\Leftrightarrow x=0\)

\(C=\dfrac{\sqrt{x}+2022}{\sqrt{x}+2022}-\dfrac{1}{\sqrt{x}+2022}=1-\dfrac{1}{\sqrt{x}+2022}\)

Do \(\sqrt{x}+2022\ge2022\Leftrightarrow\dfrac{1}{\sqrt{x}+2022}\le\dfrac{1}{2022}\Leftrightarrow-\dfrac{1}{\sqrt{x}+2022}\ge-\dfrac{1}{2022}\)

\(\Leftrightarrow C=1-\dfrac{1}{\sqrt{x}+2022}\ge1-\dfrac{1}{2022}=\dfrac{2011}{2022}\)

Dấu"=" xảy ra \(\Leftrightarrow x=0\)

Đặt A=|x-2022|+|x-2023|+|x-2024|

TH1: x<2022

=>x-2022<0; x-2023<0; x-2024<0

=>A=-x+2022-x+2023-x+2024=-3x+6069

Vì hàm số A=-3x+6069 là hàm số nghịch biến trên R

nên A nhỏ nhất khi x lớn nhất

Khi x<2022 thì x không có giá trị lớn nhất

=>A không có giá trị nhỏ nhất(1)

TH2: 2022<=x<2023

=>x-2022>=0; x-2023<0; x-2024<0

=>A=x-2022+2023-x+2024-x=-x+2025

Vì hàm số A=-x+2025 là hàm số nghịch biến trên R

nên A nhỏ nhất khi x lớn nhất

Khi 2022<=x<2023 thì x không có giá trị lớn nhất

=>A không có giá trị nhỏ nhất(2)

TH3: 2023<=x<2024

=>x-2022>0; x-2023>=0; x-2024<0

=>A=x-2022+x-2023+2024-x=x-2021

Vì hàm số A=x-2021 là hàm số đồng biến trên R

nên A nhỏ nhất khi x nhỏ nhất

Khi 2023<=x<2024 thì \(x_{\min}=2023\)

=>A min=2023-2021=2(3)

TH4: x>=2024

=>x-2022>0; x-2023>0; x-2024>=0

=>A=x-2022+x-2023+x-2024=3x-6069

Vì hàm số A=3x-6069 là hàm số đồng biến trên R

nên A nhỏ nhất khi x nhỏ nhất

Khi x>=2024 thì \(x_{\min}=2024\)

=>\(A_{\min}=3\cdot2024-6069=6072-6069=3\) (4)

Từ (1),(2),(3),(4) suy ra \(A_{\min}=3\) khi x=2023

Ta có: \(P=\frac{|x-2022|+|x-2023|+|x-2024|+2022}{|x-2022|+|x-2023|+|x-2024|}\)

\(=1+\frac{2022}{|x-2022|+|x-2023|+|x-2024|}=1+\frac{2022}{A}\)

\(A\ge3\forall x\)

=>\(\frac{2022}{A}\le\frac{2022}{3}=674\forall x\)

=>\(1+\frac{2022}{A}\le1+674=675\forall x\)

=>P<=675∀x

Dấu '=' xảy ra khi x=2023

Lời giải:

Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$

$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$

$\Rightarrow x=y=z$.

Do đó:

$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$

Lời giải:

Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$

$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$

$\Rightarrow x=y=z$.

Do đó:

$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$

a)

`(2x-1)(x+2/3)=0`

\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b)

\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)

\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)

\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)

\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)

\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)

sai rồi , x không thể có 2 giá trị

\(\dfrac{2}{3}-\left|\dfrac{3}{4}\right|+\sqrt{\dfrac{25}{9}}-\left(\dfrac{2021}{2022}\right)^0=\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{3}-1=\dfrac{7}{12}\)

\(=\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{3}-1=\dfrac{7}{12}\)

\(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)

Vì \(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)

\(\Rightarrow\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}+4=0\)

\(\Rightarrow\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)+\left(\dfrac{x+21}{2023}+1\right)+\left(\dfrac{x+20}{2024}+1\right)=0\)

\(\Rightarrow\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}+\dfrac{x+2044}{2023}+\dfrac{x+2044}{2024}=0\)

\(\Rightarrow\left(x+2044\right)\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\right)=0\)

\(\Rightarrow x+2044=0\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\ne0\right)\)

\(\Rightarrow x=-2024\)