Ta có A = 2019.2021.a = (2020 – 1)(2020 + 1)a = ( 2020 2 – 1)a

Và B   =   ( 2019 2   +   2 . 2019   +   1 ) a   =   ( 2019   +   1 ) 2 a   =   2020 2 a

Vì 2020 2   –   1   <   2020 2 và a > 0 nên ( 2020 2   –   1 ) a   <   2020 2 a hay A < B

Đáp án cần chọn là: D

a)

\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)

      =\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)

      \(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)

Vay P=a+1

phan b,c ap dung phan a la ra

CM bài toán phụ: \(x+y+z=0\) 

CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương

Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)

\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)

\(Q=2021-\frac{1}{2021}=...\)

Phần b mượn bài Upin ta có:

\(P=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

\(P=2020+1\)

\(P=2021\)

c) Áp dụng công thức \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\),ta được:

\(Q=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)

\(=1+1+1+...+1-\frac{1}{2021}\)

\(=2021-\frac{1}{2021}=\frac{4084440}{2021}\)

=>a^2+b^2+c^2+3-2a-2b-2c=0

=>(a^2-2a+1)+(b^2-2b+1)+(c62-2c+1)=0

=>(3 hằng dẳng thức của a-1 b-1 c-1)

Suy ra (a-1)^2=0

và (b-1)^2=0

và(c-1)^2=0

thay vào A suy ra A=0

cố gắng trình bày lại nhé bạn!

\(a+b+c=2020\Rightarrow\frac{1}{a+b+c}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a\left(ab+ac\right)+abc-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a^2\left(b+c\right)=0\)

\(\Leftrightarrow\left(ab+bc+ac+a^2\right)\left(b+c\right)=0\)

\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

Nếu a + b = 0 thì c = 2020

Nếu b + c = 0 thì a = 2020

Nếu a + c = 0 thì b = 2020

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Rightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)

\(\Rightarrow a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2=abc\)

\(\Rightarrow...\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(TH1:a=-b\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a}-\frac{1}{a}+\frac{1}{c}=\frac{1}{c}\)

Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2020}\Rightarrow\frac{1}{c}=\frac{1}{2020}\Leftrightarrow c=2020\)

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