S = 2 + 2\(^2\) + 2\(^3\) + ...+ 2\(^{100}\)

Xét dãy số: 1; 2; 3; 4;...;100. Dãy số này có 100 hạng tử

Vậy S + 5 có: 100 + 1 = 101(hạng tử)

Vì: 101 : 3 = 33 (dư 2) nên nhóm ba hạng tử liên tiếp của S + 5 thành một nhó thì khi đó:

S + 5 = (2 + 5) + (2\(^2\) + 2\(^3\) + 2\(^4\))+ ....+ (2\(^{98}\) + 2\(^{99}\) + 2\(^{100}\))

S + 5 = 7 + 2\(^2\).(1 + 2 + 2\(^2\)) + ...+ 2\(^{98}\).(1 + 2 + 2\(^2\))

S + 5 = 7+ 2\(^2\) .7 + ...+ 2\(^{98}\).7

S + 5 = 7.(1 + 2\(^2\) + ...+ 2\(^{98}\)) ⋮ 7 (đpcm)

\(\dfrac{5}{9}+\dfrac{1}{3}:x=\dfrac{2}{3}\\ \dfrac{1}{3}:x=\dfrac{2}{3}-\dfrac{5}{9}=\dfrac{1}{9}\\ x=\dfrac{1}{3}:\dfrac{1}{9}=3\)

\(\dfrac{5}{9}+\dfrac{1}{3}:x=\dfrac{2}{3}\)

=>\(\dfrac{1}{3}:x=\dfrac{2}{3}-\dfrac{5}{9}=\dfrac{1}{9}\)

=>\(x=\dfrac{1}{3}:\dfrac{1}{9}=\dfrac{9}{3}=3\)

\(\left(x+\dfrac{3}{4}\right)^2=\dfrac{1}{4}\)

=>\(\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{2}\\x+\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{1}{4}\\x=-\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{5}{4}\end{matrix}\right.\)

\(\frac{x-1}{2}\) = \(\frac{8}{x-1}\)

(\(x-1\)).(\(x-1\)) = 8.2

(\(x\) - 1)\(^2\) = 16

(\(x-1\))\(^2\) = 4\(^2\)

\(\left[\begin{array}{l}x-1=-4\\ x-1=4\end{array}\right.\)

\(\left[\begin{array}{l}x=-4+1\\ x=4+1\end{array}\right.\)

\(\left[\begin{array}{l}x=-3\\ x=5\end{array}\right.\)

Vậy \(x\in\) {-3; 5}

\(\dfrac{x-1}{2}=\dfrac{8}{x-1}\)

=>\(\left(x-1\right)\left(x-1\right)=2\cdot8=16\)

=>\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

x=33/11:6

x=1/2

❤❤❤

\(\frac{6}{x}\) = - \(\frac{33}{11}\)

\(\frac{6}{x}\) = - 3

\(x\) = 6 : -3

\(x=-2\)

Vậy \(x=-2\)

xy+3x-7y=21

x.(y+3)-7y-21=21-21

x.(y+3)-7.(y+3)=0

(x-7).(y+3)=0

=>x-7, y+3 ∈ 0

=>x=-7 , y= -3

Vậy ( x, y) ∈ ( -7, -3)

\(2^{3n}=\left(2^3\right)^n=8^n\)

\(3^{2n}=\left(3^2\right)^n=9^n\)

mà \(8^n< 9^n\left(8< 9\right)\)

nên \(2^{3n}< 3^{2n}\)

\(-\frac{11}{12}<\frac{x}{5}<-\frac{11}{15}\)

\(-11.5<\frac{12x}{5.12}<-\frac{11.4}{15.4}\)

\(-\frac{55}{60}<\frac{12x}{60}<-\frac{44}{60}\)

\(-55<12x<-44\)

\(-\frac{55}{12}

\(\rArr x=-4\left(x\in Z\right)\)

Do n chia 3;5;7 lần lượt dư 2;4;6

\(\Rightarrow n+1\) chia hết cho cả 3;5 và 7

\(\Rightarrow n+1=ƯC\left(3;5;7\right)=Ư\left(105\right)\) 

Mà n nhỏ nhất \(\Rightarrow n+1\) nhỏ nhất

\(\Rightarrow n+1=105\)

\(\Rightarrow n=104\)

Gọi mẫu là x

Theo đề, ta có: \(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)

=>\(\dfrac{-55}{60}< \dfrac{-55}{-11x}< \dfrac{-55}{75}\)

=>\(\dfrac{55}{60}>\dfrac{55}{-11x}>\dfrac{55}{75}\)

=>60<-11x<75

=>\(-\dfrac{60}{11}>x>-\dfrac{75}{11}\)

=>\(x=-6\)

Vậy: Phân số cần tìm là \(\dfrac{5}{-6}\)