a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

\(=\left(x+1\right)^2\)

\(=x^2+2x+1\)

a: \(\Leftrightarrow x-3=7\)

hay x=10

x⁴ + 2021x² - 2020x + 2021

= (x⁴ + x) + 2021(x² - x + 1)

= x(x³ + 1) + 2021(x² - x + 1)

= x(x + 1)(x² - x + 1) + 2021(x² - x + 1)

= (x² + x + 2021)(x² - x + 1)

Ta có: \(\left|x+\frac{1}{2021}\right|\ge0\) ; \(\left|x+\frac{2}{2021}\right|\ge0\) ; ... ; \(\left|x+\frac{2020}{2021}\right|\ge0\) \(\left(\forall x\right)\)

\(\Rightarrow\left|x+\frac{1}{2021}\right|+\left|x+\frac{2}{2021}\right|+...+\left|x+\frac{2020}{2021}\right|\ge0\left(\forall x\right)\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

Từ đó ta được: \(x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Leftrightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Leftrightarrow x=\frac{\left(2020+1\right)\left[\left(2020-1\right)\div1+1\right]}{2021}\)

\(\Leftrightarrow x=\frac{2021\cdot2020}{2021}=2020\)

Vậy x = 2020

\(\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|=2021x\)

Ta có:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)

\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Rightarrow x=2020\)

Ta có : \(\left(2020.x^2+2021\right).\left(x^2-1\right).\left(2.x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2020.x^2+2021=0\\x^2-1=0\\2.x+=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\notinℝ\\x=\pm1\\x=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=-1\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x=\left\{\pm1;-\frac{1}{2}\right\}\)

\(a,Sửa:2021x-1+2022x\left(1-2021x\right)=0\\ \Leftrightarrow\left(2021x-1\right)\left(1-2022x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2021}\\x=\dfrac{1}{2022}\end{matrix}\right.\)

ĐKXĐ: -1/2021<=x<=1/2021

TH1: x=0

Khi x=0 thì ta có;

\(VT=\sqrt{1+2020\cdot0}+\sqrt{1-2020\cdot0}=1+1=2\)

\(VP=\left(1+2021\cdot0\right)\cdot\sqrt{1-2021\cdot0}+\left(1-2021\cdot0\right)\cdot\sqrt{1+2021\cdot0}\)

=1+1

=2

=>VT=VP

=>x=0 là nghiệm của phương trình

TH2: x<>0

\(\text{VT}^2 = (\sqrt{1 + 2020x} + \sqrt{1 - 2020x})^2 \le (1^2 + 1^2)(1 + 2020x + 1 - 2020x) = 2 \cdot 2 = 4\)

Dấu '=' xảy ra khi \(\sqrt{1 + 2020x} = \sqrt{1 - 2020x} \Leftrightarrow x = 0\) (loại)

=>VT>2

\(\text{VP} = \sqrt{1 + 2021x} \cdot \sqrt{1 - 2021x} \cdot \left(\sqrt{1 + 2021x} + \sqrt{1 - 2021x}\right) = \sqrt{1 - 2021^2x^2} \cdot \left(\sqrt{1 + 2021x} + \sqrt{1 - 2021x}\right)\)

Khi 0<x<=1/2021 thì ta sẽ có:

2021x+1>2020x+1; 1-2021x<1-2020x

Xét hàm số \(f(a)=(1+a)\sqrt{1-a}+(1-a)\sqrt{1+a}\)

=>\(f(a) = \sqrt{1-a^2}(\sqrt{1+a} + \sqrt{1-a})\)

Đặt \(\sin\alpha=2021x\) , với α∈(0;pi/2]

\(\text{VP} = (1 + \sin\alpha)\sqrt{1 - \sin\alpha} + (1 - \sin\alpha)\sqrt{1 + \sin\alpha}\)

\(=\left(1+\sin\alpha\right)\left(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\right)+\left(1-\sin\alpha\right)\left(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}\right)\)

\(=2\cos\frac{\alpha}{2}-2\sin\alpha\sin\frac{\alpha}{2}=2\cos\frac{\alpha}{2}-4\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\sin\frac{\alpha}{2}=2\cos\frac{\alpha}{2}\left(1-2\sin^2\frac{\alpha}{2}\right)=2\cos\frac{\alpha}{2}\cos\alpha\)

Ta thấy: f(x) đạt cực đại bằng 2 tại x = 0 và giảm dần (hàm nghiêm ngặt giảm) khi |x| tăng

mà tốc độ giảm của VP khi x > 0 nhanh hơn VT, hoặc xét hàm đặc trưng \(g(t) = \sqrt{1+tx} + \sqrt{1-tx}\) giảm theo t.

Do đó: Khi x<>0 thì VT<>VP

Vậy: x=0

Lời giải:

$x(x-1)+2021-2021x=0$

$\Leftrightarrow x(x-1)-(2021x-2021)=0$

$\Leftrightarrow x(x-1)-2021(x-1)=0$

$\Leftrightarrow (x-1)(x-2021)=0$

$\Leftrightarrow x-1=0$ hoặc $x-2021=0$

$\Leftrightarrow x=1$ hoặc $x=2021$