:)))) :>>> :+ := :- :

185/741

ta co tu suy nghi 

chuẩn CMNR ;+

a) \(A=2^{100}-2^{99}-2^{98}-...-2^2-2^1\)( Có 2 câu nên mình tính nhanh luôn nhé )

\(\Leftrightarrow A=2^{100}-\left(2^1+2^2+2^3+...+2^{98}+2^{99}\right)\)

\(A=2^{100}-\left(2^{100}-2^1\right)=2^{100}-2^{100}+2=2\)

b) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{36.37.38}+\frac{1}{37.38.39}\)

\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{38-36}{36.37.38}+\frac{39-37}{37.38.39}\)

\(=\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}\right)+\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}\right)+...+\left(\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)

\(=\left(\frac{1}{2}-\frac{2}{3}\right)+\left(\frac{2}{3}-\frac{3}{4}\right)+\left(\frac{3}{4}-\frac{4}{5}\right)+...+\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}-\frac{2}{3}+\frac{2}{3}-\frac{3}{4}+\frac{3}{4}-\frac{4}{5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{2}-\frac{1}{38.39}=\frac{741}{1482}-\frac{1}{1482}=\frac{740}{1482}=\frac{370}{741}\)

\(\left(\frac{x+1}{39}+1\right)+\left(\frac{x+2}{38}+1\right)=\left(\frac{x+3}{37}+1\right)+\left(\frac{x+4}{36}+1\right)\)

\(\Leftrightarrow\frac{x+40}{39}+\frac{x+40}{38}-\frac{x+40}{37}-\frac{x+40}{36}=0\)

\(\Leftrightarrow\left(x+40\right)\left(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\right)=0\)

<=> x+40=0 (vì \(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\ne\)0)

<=> x=-40

Vậy x=-40

\(\dfrac{x+1}{39}+\dfrac{x+2}{38}+\dfrac{x+3}{37}=0\)

\(\Leftrightarrow\dfrac{x+1}{39}+1+\dfrac{x+2}{38}+1+\dfrac{x+3}{37}+1-3=0\)

\(\Leftrightarrow\dfrac{x+40}{39}+\dfrac{x+40}{38}+\dfrac{x+40}{37}=3\)

\(\Leftrightarrow\left(x+40\right)\left(\dfrac{1}{39}+\dfrac{1}{38}+\dfrac{1}{37}\right)=3\)

\(\Leftrightarrow\left(x+40\right).\dfrac{4331}{54834}=3\)

\(\Leftrightarrow x+40=\dfrac{164502}{4331}\)

\(\Leftrightarrow x=\dfrac{-8738}{4331}\)

-Vậy \(S=\left\{\dfrac{-8738}{4331}\right\}\)