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:

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

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

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

Áp dụng bài vừa chứng minh bên dưới :D

\(\Rightarrow P=2021\)

\(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge2\sqrt{\frac{a^2}{a^2}}+2\sqrt{\frac{b^2}{b^2}}+2\sqrt{\frac{c^2}{c^2}}=6\)

Dấu = xảy ra khi a^4=b^4=c^4=1 <=> \(a=\pm1;b=\pm1;c\pm1\)

-> B = 3

gọi a/2019=b/2020=c/2021 là x

\(\Rightarrow\)a=2019*x ;b=2020*x;c=2021*x

\(\Rightarrow\)M=4*(2019*x-2020*x)*(2020-2021)-(2021*x-2019*x)^2

\(\Rightarrow\)M=4*(-x)*(-x)-(2x)^2

\(\Rightarrow\)M=4*x^2-4*x^2

⇒M=0

Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)

Khi đó M = 4(a - b)(b - c) - (c - a)² 

= 4(2020k - 2021k)(2021k - 2022k) - (2022k - 2020k)²

= 4(-k)(-k) - (2k)²

= 4k² - 4k² = 0

Vậy M = 0

Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\)( \(k\ne0\))

\(\Rightarrow a=2020k\); \(b=2021k\); \(c=2022k\)

Thay a, b, c vào biểu thức M ta có:

\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)

     \(=4\left(2020k-2021k\right)\left(2021k-2022k\right)-\left(2022k-2020k\right)^2\)

      \(=4.\left(-k\right).\left(-k\right)-\left(2k\right)^2=4k^2-4k^2=0\)

Vậy \(M=0\)

a) A = \(\frac{a+1-2\sqrt{a}}{a+1}:\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\right)\)

= \(\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\)

= \(\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2}\) = \(\sqrt{a}+1\)

b) a = \(2020-2.\sqrt{2019}\) = \(\left(\sqrt{2019}-1\right)^2\)

=> \(A=\sqrt{\left(\sqrt{2019}-1\right)^2}+1\) = \(\sqrt{2019}\)

nếu đó là câu giải pt thì bắt buộc phải đặt ĐKXĐ, nếu đó là câu rút gọn thì không cần nhé

Xét phân thức phụ sau, với n nguyên dương lớn hơn 1 ta có:

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

\(< \frac{2\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}\right)^2\sqrt{n}}=2\left(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}\right)\sqrt{n}}\right)\)

\(=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

=> \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Áp dụng vào bài toán ta được:

\(A=2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\right)\)

\(A=2-\frac{2}{\sqrt{2020}}< 2=B\)

Vậy A < B

A.2

......

Chúc học tốt