ta có 

\(C=2020\times\left(2021^9+2021^8+...+2021^2+2021^1+1\right)+1\)

\(2020\times\frac{2021^{10}-1}{2021-1}+1=2021^{10}-1+1=2021^{10}\)

\(\dfrac{5^{2021}}{5^{2020}}\cdot5^2=5\cdot5^2=5^3\)

5²⁰²¹:5²⁰²⁰. 5²

= 5¹.5²

=5³

5^2021 : 5^2020 x 5^2

=  5^1 x 5^2

= 5^3

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:

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

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

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

\(\sqrt{\left(1-\sqrt{2020}\right)^2\cdot\left(\sqrt{2021-2\sqrt{2020}}\right)}\)

\(=\sqrt{\left(1-\sqrt{2020}\right)^2\cdot\sqrt{\left(1-\sqrt{2020}\right)^2}}\)

\(=\sqrt{\left(1-\sqrt{2020}\right)^2\cdot\left(1-\sqrt{2020}\right)}\)

\(=\sqrt{\left(1-\sqrt{2020}\right)^3}=\left(1-\sqrt{2020}\right)\sqrt{1-\sqrt{2020}}\)

\(E=\left(a-1\right)\left(x^2+1\right)-x\left(y+1\right)+\left(x+y^2-a+1\right)\)

\(=x^2a+a-x^2-1-xy-x+x+y^2-a+1\)

\(=x^2a-x^2-xy+y^2\)

Vậy...

\(A=\left(\dfrac{2020}{2021}xy^5z\right).\left(\dfrac{2020}{2021}x^3yz^2\right).\left(-\dfrac{2020}{2021}\right)^0\)

\(a)A=\dfrac{2020.2021.2020}{2021.2020.2021}.\left(x.x^3\right).\left(y^5.y\right).\left(z.z^2\right)\Leftrightarrow A=\dfrac{2020}{2021}x^4.y^6.z^3\)

\(b)A=\dfrac{2020}{2021}x^4.y^6.z^3\)

\(\Rightarrow\text{A có hệ số là:}\dfrac{2020}{2021}\)

\(\text{Phần biến là:}\left(x,y,z\right)\)

\(c)\text{Xét A ta có:}\dfrac{2020}{2021}< 0;x^4,y^6\text{ luôn }< 0\)

\(\Rightarrow\dfrac{2020}{2021}x^4.y^6>0\Rightarrow\text{ Nếu }z< 0\Rightarrow A\le0\text{ và z có số mũ là:3}\)

\(\text{Chẳng hạn:}\left(-\right).\left(-\right).\left(-\right)=\left(-\right).< 0\Rightarrow z\text{ phải }\ge0\text{ thì }A\ge0\)

\(\Rightarrow Z\in N\)

Ta có: \(A=\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2021}}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2021}-\sqrt{2020}\)

\(=-\sqrt{2}+\sqrt{2021}\)