Hướng làm:

Thấy cả tử mẫu cộng lại đều bằng 2021 → Cộng thêm 1 rồi quy đồng với mỗi phân thức

\(\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\\ \Leftrightarrow\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\\ \Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}\right)=0\\ \Leftrightarrow x+2021=0\Leftrightarrow x=-2021\)

\(< =>\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\)

\(< =>\dfrac{x+2+2019}{2019}+\dfrac{x+3+2018}{2018}=\dfrac{x+4+2017}{2017}+\dfrac{x+2021}{2021}\)

\(< =>\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\)

\(< =>\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}=\right)=0\)

\(< =>x+2021=0< =>x=-2021\)

Vậy....

\(\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x}{2021}\)

\(\Leftrightarrow\frac{x+2}{2019}+1+\frac{x+3}{2018}+1=\frac{x+4}{2017}+1+\frac{x}{2021}+1\)

\(\Leftrightarrow\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2021}\)

\(\Leftrightarrow x+2021=0\)

\(\Leftrightarrow x=-2021\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a: Thay x=-3 vào B, ta được:

\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)

b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)

\(x^{2019}-y^{2019}+2\left(x-y\right)=0\)

<=> \(\left(x-y\right)\left(x^{2018}+x^{2017}y+...+xy^{2017}+y^{2018}\right)+2\left(x-y\right)=0\)

<=> \(\left(x-y\right)\left(x^{2018}+x^{2017}y+...+xy^{2017}+y^{2018}+2\right)=0\)(1)

Có: \(x^{2018}+x^{2017}y+...+xy^{2017}+y^{2018}+2>0\)mọi x, y.

(1) <=> \(x-y=0\)

<=> x = y

Thế vào P ta có:

\(P=x^4-2x^2+2=\left(x^2-1\right)^2+1\ge1\)

"=" xảy ra <=> \(y=x=\pm1\)

Vậy min P =1 khi và chỉ khi x = y =1 hoặc x = y =-1.

  2021 x 2020 + 21 + 2000 - 2021 x 2020

= 2021 x (2020-2020) +21 +2000

= 2021 x 0 + 21 + 2000

= 0 + 21 + 2000

= 2021

Cách này làm nhanh hơn cách của em nè. Chúc em học tốt nha <3

e cảm ơn ạ

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10