C = A - B

= (x - 3x³ + 1 + 4x²) - (x - x³ - 2022 - 2x³ - 2x²)

= x - 3x³ + 1 + 4x² - x + x³ + 2022 + 2x³ + 2x²

= (-3x³ + x³ + 2x³) + (4x² + 2x²) + (1 + 2022)

= 6x² + 2023

Do x² ≥ 0 với mọi x

⇒ 6x² ≥ 0 với mọi x

⇒ 6x² + 2023 > 0 với mọi x

Vậy C luôn dương với mọi x

C = A - B

= (x - 3x³ + 1 + 4x²) - (x - x³ - 2022 - 2x³ - 2x²)

= x - 3x³ + 1 + 4x² - x + x³ + 2022 + 2x³ + 2x²

= (-3x³ + x³ + 2x³) + (4x² + 2x²) + (1 + 2022)

= 6x² + 2023

Do x² ≥ 0 với mọi x

⇒ 6x² ≥ 0 với mọi x

⇒ 6x² + 2023 > 0 với mọi x

Vậy C luôn dương với mọi x

\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)

\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(c,\\ \left(2x+1\right)^3=9.81\\ \left(2x+1\right)^3=3^2.3^4\\ \left(2x+1\right)^3=3^6\\ \left(2x+1\right)^3=\left(3^2\right)^3=9^3\\ Vậy:2x+1=9\\ 2x=9-1\\ 2x=8\\ x=\dfrac{8}{2}\\ x=4\)

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

a) + Chia thành 2 trường hợp 

- 2x - 1 = 0

2x = 0 + 1

2x = 1

x = 1 : 2

x = 0,5

- x + 2/3 = 0

x = 0 - 2/3

x = -2/3

vậy x = { 0,5 ; -2/3 }

\(A=2022+\sqrt{2\left(x^2-2x+1\right)+1}\\ A=2022+\sqrt{2\left(x-1\right)^2+1}\ge2022+\sqrt{1}=2023\\ A_{min}=2023\Leftrightarrow x=1\)

\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)

\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)

\(\Rightarrow A< B\)

a: (2x-3)^2>=0

=>-(2x-3)^2<=0

=>D<=-3

Dấu = xảy ra khi x=3/2

b: (2x-5)^2>=0

(y+1/2)^2>=0

=>(2x-5)^2+(y+1/2)^2>=0

=>D>=2022

Dấu = xảy ra khi x=5/2 và y=-1/2

P = (x^2 + 2x) - 2024
= (x^2 + 2x + 1) - 1 - 2024
= (x + 1)^2 - 2025

Với mọi giá trị của x, (x + 1)^2 luôn lớn hơn hoặc bằng 0. Do đó, giá trị nhỏ nhất của P là khi (x + 1)^2 đạt giá trị nhỏ nhất, tức là bằng 0.

Khi (x + 1)^2 = 0, ta có x + 1 = 0, từ đó suy ra x = -1.

Vậy, giá trị nhỏ nhất của biểu thức P là P = (-1 + 1)^2 - 2025 = -2025.

\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)