Câu 1:

\(C=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\) 

\(C=\dfrac{2}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\) 

\(C=\dfrac{2}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\) 

\(C=\dfrac{2}{3}.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\) 

\(C=\dfrac{2}{3}.\dfrac{99}{100}\) 

\(C=\dfrac{33}{50}\)

Câu 3:

a) Gọi ƯCLN(2n+5;n+3)=d

\(\Rightarrow\left\{{}\begin{matrix}2n+5⋮d\\n+3⋮d\end{matrix}\right.\)           \(\Rightarrow\left\{{}\begin{matrix}2n+5⋮d\\2.\left(n+3\right)⋮d\end{matrix}\right.\)        \(\Rightarrow\left\{{}\begin{matrix}2n+5⋮d\\2n+6⋮d\end{matrix}\right.\) 

\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\) 

\(\Rightarrow1⋮d\) 

\(\Rightarrow d=1\) 

Vậy \(\dfrac{2n+5}{n+3}\) là p/s tối giản

b) Để \(B=\dfrac{2n+5}{n+3}\) là số nguyên thì \(2n+5⋮n+3\) 

\(2n+5⋮n+3\) 

\(\Rightarrow2n+6-1⋮n+3\) 

\(\Rightarrow1⋮n+3\) 

\(\Rightarrow n+3\inƯ\left(1\right)=\left\{-1;1\right\}\) 

Ta có bảng giá trị:

\(n+3=-1\rightarrow n=-4\) 

\(n+3=1\rightarrow n=-2\) 

Vậy \(n\in\left\{-4;-2\right\}\)

Câu 2:

a) \(M=5+5^2+5^3+...+5^{80}\) 

\(M=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^{79}.\left(1+5\right)\) 

\(M=\left(1+5\right).\left(5+5^3+...+5^{79}\right)\) 

\(M=6.\left(5+5^3+...+5^{79}\right)⋮6\left(đpcm\right)\) 

b) Ta thấy:

\(M=5+5^2+5^3+...+5^{80}⋮5\)  

Mặt khác, do:

 \(5^2+5^3+...+5^{80}⋮5^2\) (vì tất cả các số hạng này đều chia hết cho 5²)

\(\Rightarrow M=5+5^2+5^3+...+5^{80}⋮̸5^2\) (do 5 \(⋮̸\) 5²)

\(\Rightarrow M⋮5\) nhưng \(M\) \(⋮̸̸\) \(5^2\) 

\(\Rightarrow M\) không phải là số chính phương (đpcm)

1.A

2.A

3.A

4.C

Câu 1 khoanh A

Câu 2 khoanh A

Câu 3 khoanh A

Câu 4 khoanh C

B - A - A

1.B

2.A

3.A

b

a

a

chữ m quá 

ll

1. đúng rồi

2. ofre

3.he=> the

\(x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\\ x+\dfrac{3}{5}=\dfrac{4}{25}\\ x=\dfrac{4}{25}-\dfrac{3}{5}\\ x=\dfrac{4}{25}-\dfrac{15}{25}\\ x=-\dfrac{11}{25}\)

__

\(\left|x+\dfrac{3}{4}\right|-\dfrac{5}{6}=0\\ \left|x+\dfrac{3}{4}\right|=0+\dfrac{5}{6}\\ \left|x+\dfrac{3}{4}\right|=\dfrac{5}{6}\\ \left|x+\dfrac{3}{4}\right|=\pm\dfrac{5}{6}\\ \left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{5}{6}\\x+\dfrac{3}{4}=-\dfrac{5}{6}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{6}-\dfrac{3}{4}\\x=-\dfrac{5}{6}-\dfrac{3}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{20}{24}-\dfrac{18}{24}\\x=-\dfrac{20}{24}-\dfrac{18}{24}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{2}{24}\\x=-\dfrac{38}{24}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{12}\\x=-\dfrac{19}{12}\end{matrix}\right.\)

__

\(\left(x+\dfrac{3}{7}\right)^2=\dfrac{25}{49}\\ \left(x+\dfrac{3}{7}\right)^2=\left(\pm\dfrac{5}{7}\right)^2\\ \left[{}\begin{matrix}x+\dfrac{3}{7}=\dfrac{5}{7}\\x+\dfrac{3}{7}=-\dfrac{5}{7}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{7}-\dfrac{3}{7}\\x=-\dfrac{5}{7}-\dfrac{3}{7}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{2}{7}\\x=-\dfrac{8}{7}\end{matrix}\right.\)

Câu 2: 

a: x=4/25-3/5=4/25-15/25=-11/25

b: =>|x+3/4|=5/6

=>x+3/4=5/6 hoặc x+3/4=-5/6

=>x=5/6-3/4=10/12-9/12=1/12 hoặc x=-10/12-9/12=-19/12

c: =>x+3/7=5/7 hoặc x+3/7=-5/7

=>x=-8/7 hoặc x=2/7