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

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

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:

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

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

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

TA XÉT PHÂN THỨC TỔNG QUÁT SAU:   

\(A=\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}\)

\(A=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)

\(A=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)

\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)

\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)

\(A=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

THAY LẦN LƯỢT CÁC GIÁ TRỊ n từ 1 => 2021 vào ta được: 

=>    \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)

=>   \(A=1-\frac{1}{\sqrt{2021}}=\frac{\sqrt{2021}-1}{\sqrt{2021}}\)

VẬY    \(A=\frac{\sqrt{2021}-1}{\sqrt{2021}}.\)

Ta có: \(\frac{1}{\left(a-1\right)\sqrt{a}+a.\sqrt{a-1}}=\frac{a-\left(a-1\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}\)

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

\(=\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a-1}}-\frac{\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}=\frac{1}{\sqrt{a-1}}-\frac{1}{\sqrt{a}}\)

Thay lần lượt các giá trị của a bằng \(2;3;4;........;2021\)ta được:

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

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

Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)

\(\Rightarrow1+2020^2=2021^2-2.2020\)

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

\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)

Đk: \(\forall x\in R\)

Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=2021\)

Lập bảng xét dầu

x                   -2                   1 

x - 1   -         |           -          0       +

x + 2   -        0         +          |            -

Xét các TH xảy ra :

TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021

<=> -2x = 2022 <=> x = -1011 (tm)

TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021

<=> 0x = 2018 (vô lí) => pt vô nghiệm

TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021

<=> 2x = 2020 <=> x = 1010 (tm)

Vậy S = {-1011; 1010}

\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+n^2+\left(n+1\right)^2}{n^2\left(n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(n^2+n\right)^2+n^2+n^2+2n+1}{\left(n^2+n\right)^2}}=\sqrt{\dfrac{\left(n^2+n\right)^2+2\left(n^2+n\right)+1}{\left(n^2+n\right)^2}}\)

\(=\sqrt{\dfrac{\left(n^2+n+1\right)^2}{\left(n^2+n\right)^2}}=\dfrac{n^2+n+1}{n^2+n}=1+\dfrac{1}{n\left(n+1\right)}\)

\(\Rightarrow A=1+\dfrac{1}{2.3}+1+\dfrac{1}{3.4}+....+1+\dfrac{1}{2021.2022}\)

\(=2020+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

\(=2020+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=2020+\dfrac{1}{2}-\dfrac{1}{2022}=...\)

\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}=\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{3}}=\sqrt{\left(1+\dfrac{1}{2}-\dfrac{1}{3}\right)^2}=1+\dfrac{1}{2}-\dfrac{1}{3}\)

Cmttt ta được:

\(A=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}+1+\dfrac{1}{2021}-\dfrac{1}{2022}\\ A=2020+\dfrac{1}{2}-\dfrac{1}{2022}=2020+\dfrac{505}{1011}=...\)

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