X nó là số nguyên nhưng nó nhỏ hơn x í mà

\(\Rightarrow A=1+1+1+1+1+1+1+1+1+1=10_{ }\)

\(\left|3,5-x\right|>=0\forall x\)

=>\(\left|3,5-x\right|+1,9>=1,9\forall x\)

=>B>=1,9 với mọi x

Dấu '=' xảy ra khi x=3,5

vậy: \(B_{min}=1,9\) khi x=3,5

\(\left\lbrack\left(2,1^2-4,01\right)^2\cdot2,5+7,3\right\rbrack:\left\lbrack1,9-0,4^2-\left(\frac45\right)^2\right\rbrack\)

\(=\left\lbrack\left(4.41-4,01\right)^2\cdot2,5+7,3\right\rbrack:\left\lbrack1,9-0,16-0,64\right\rbrack\)

\(=\left\lbrack0,4^2\cdot2,5+7,3\right\rbrack:\left\lbrack1,9-0,8\right\rbrack=7,7:1,1=7\)

a/\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)
\(=\dfrac{1}{3}\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{1018}{1019}\)
\(=\dfrac{1018}{3057}\)
b/\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)
\(=\dfrac{7}{8}\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{2018}{2019}\)
\(=\dfrac{7063}{8076}\)

Thankyou, Love you

a: \(\frac12\cdot2^{n}+4\cdot2^{n}=9\cdot5^{n}\)

=>\(\frac92\cdot2^{n}=9\cdot5^{n}\)

=>\(\left(\frac25\right)^{n}=9:\frac92=2\)

=>\(n=\log_{0,4}2\)

b: \(2^{n}\left(\frac12+4\right)=9\cdot5^{n}\)

=>\(2^{n}\cdot\frac92=9\cdot5^{n}\)

=>\(\left(\frac25\right)^{n}=9:\frac92=2\)

=>\(n=\log_{0.4}2\)

c: \(2^{n-1}\cdot9=9\cdot5^{n}\)

=>\(2^{n}\cdot\frac92=9\cdot5^{n}\)

=>\(\left(\frac25\right)^{n}=9:\frac92=2\)

=>\(n=\log_{0.4}2\)

A=( -4,3 . 1,1 + 1,1 . 4,5 ) : ( -0,5 : 0,05 + 10,01 )

A= [1,1.(-4.3+4,5)] : (-10+10,01)

A=(1,1 . 0,2): 0,01

A=0,22 : 0,01

A=22

\(\left(-4,3.1,1+1,1.4,5\right):\left(-0,5:0,05+10,01\right)\)

\(=\left[1,1\left(-4,3+4,5\right)\right]:\left(-10+10,01\right)\)

\(=\left[1,1.0,2\right]:0,01\)

\(=0,22:0,01\)

\(=22\)