B=/x/+2021 / 2020
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Đặt \(2020-x=u;x-2021=v\)thì \(u+v=-1\)
Phương trình trở thành \(\frac{u^2+uv+v^2}{u^2-uv+v^2}=\frac{19}{49}\Leftrightarrow30u^2+30v^2+68uv=0\)
\(\Leftrightarrow15\left(u+v\right)^2+4uv=0\Leftrightarrow4uv=-15\Leftrightarrow uv=\frac{-15}{4}\)
hay \(\left(2020-x\right)\left(x-2021\right)=-\frac{15}{4}\Leftrightarrow x^2-4041x+4082416,25=0\)
Dùng công thức nghiệm tìm được x = 2022, 5 hoặc x = 2018, 5
B=2021 x 13 + 2009 + 2020 x 2007/2020 + 2020 x520x2020
B=2021 x 13 +2009+2020/1x2007/2020 +2020x520x2020
B=26273+2009+20201/1x2007/20201+2121808000
B=28282+2007+2121808000
B=2121838289
Ta có :\(\frac{a+2020}{a-2020}=\frac{b+2021}{b-2021}\)
=> \(\frac{a+2020}{a-2020}-1=\frac{b+2021}{b-2021}-1\)
=> \(\frac{4040}{a-2020}=\frac{4042}{b-2021}\)
=> \(1:\frac{4040}{a-2020}=1:\frac{4042}{b-2021}\)
=> \(\frac{a-2020}{4040}=\frac{b-2021}{4042}\)
=> \(\frac{a-2020}{4040}+2=\frac{b-2021}{4042}+2\)
=> \(\frac{a}{4040}=\frac{b}{4042}\)
=> \(\frac{a}{2020}.\frac{1}{2}=\frac{b}{2021}.\frac{1}{2}\)
=> \(\frac{a}{2020}=\frac{b}{2021}\)(đpcm)
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: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)