a)\(M=\frac{2019\times2020-2}{2018+2018\times2020}=\frac{2019\times2020-2}{2018+2018\times2020+2020-2020}=\frac{2019\times2020-2}{\left(2018+1\right)\times2020+2018-2020}=\frac{2019\times2020-2}{2019\times2020-2}=1\\ N=\frac{-2019\times20202020}{20192019\times2020}=\frac{-2019\times10001\times2020}{2019\times10001\times2020}=-1\)

b)\(5\left|x-1\right|=3M-2N=5\\ \left|x-1\right|=1\Rightarrow\hept{\begin{cases}x-1=1\Rightarrow x=2\\x-1=-1\Rightarrow x=0\end{cases}}\)

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

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

\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)

\(=5^{2021-\dfrac{1}{2021}}\)

\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)

\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)

Bài 1:

\(A=124-5\left|x-7\right|\Leftrightarrow-5\left|x-7\right|+124\)

+Có: \(-5\left|x-7\right|\le0với\forall x\\ \Rightarrow-5\left|x-7\right|+124\le124\\ \Leftrightarrow A\le124\)

+Dấu "=" xảy ra khi \(\left|x-7\right|=0\Leftrightarrow x=7\)

+Vậy \(B_{min}=124\) khi \(x=7\)

Nếu m = 10, n = 5 , p = 2 thì

a) m + n + p = 10 + 5 + 2 = 17

m + ( n + p) = 10 + ( 5 + 2) = 17

b) m - n - p = 10 - 5 - 2 = 3

m - ( n + p )= 10 - (5 + 2) = 3

c) m + n x p = 10 + 5 x 2 = 20

(m + n) x p = ( 10 + 5 ) x 2 = 30

a,10+5+2

=17

b,10+5-2

=13

c,10+5×2

=20

a = 17

b = 13

c = 30

a)ta co:m+n+p=10+5+2=17

b)ta co:m+n-p=10+5-2=13

c)ta co:m+n*p=10+5*2=10+10=20

Thay số:

B = 2020 : 2 + 2015 + n = 4000

B = 1010 + 2015 + n = 4000

B = 3025 + n = 4000

B = n = 4000 - 3025

B = n = 975