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10 tháng 10 2021

\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

12 tháng 10 2021

a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)

\(\Rightarrow2x=-4\Rightarrow x=-2\)

b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

pls help me mk đang cần vội :(

9 tháng 11 2021

Bài 1:

\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)

Bài 2:

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)

Bài 2: 

c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

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

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

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

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

7 tháng 3 2021

Ta có x = 2020

=> x + 1 = 2021

A = x2021 - 2021x2020 + .... + 2021x - 2021

= x2021 - (x + 1)x2020 + .... + (x + 1)x - (x + 1)

= x2021 - x2021 - x2020 + .... + x2 + x - x + 1

= 1

Vậy A = 1

7 tháng 3 2021

Ta có : \(x=2020\Rightarrow x+1=2021\)

\(A=x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}-\left(x+1\right)x^{2018}+...-\left(x+1\right)x^2+\left(x+1\right)x-2021\)

= x2021 - x2021 - x2020  + x2020 + x2019 - x2019 - x2018 + ... - x3 - x2 + x+ x - 2021 = x - 2021 

mà x = 2020 hay 2020 - 2021 = -1 

Vậy với x = 2020 thì A = -1

30 tháng 8 2021

\(2021x\left(x-2020\right)-x+2020=0\)

\(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

Ta có: \(2021x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

16 tháng 3 2023

A = \(\dfrac{x^2-2x+2020}{2021x^2}\)

\(\dfrac{2020x^2-2.2020.x+2020^2}{2021.2020x^2}\)

\(=\dfrac{2019x^2}{2021.2020x^2}+\dfrac{x^2-2.2020.x+2020^2}{2021.2020x^2}\)

\(\dfrac{2019}{2021.2020}+\dfrac{\left(x-2020\right)^2}{2021.2020x^2}\ge\dfrac{2019}{2021.2020}\)

Dấu "=" xảy ra <=> x - 2020 = 0

                       <=> x = 2020

Vậy minA = \(\dfrac{2019}{2021.2020}\)đạt được tại x = 2020