a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)

\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)

b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(=\frac{15}{16}\)

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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\(a)\) \(A=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{49.51}\)

\(A=3\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.50}\right)\)

\(2A=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.50}\right)\)

\(2A-A=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=3\left(\frac{1}{3}-\frac{1}{50}\right)\)

\(A=1-\frac{3}{50}\)

\(A=\frac{47}{50}\)

Vậy \(A=\frac{47}{50}\)

\(b)\) \(B=\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(B=1-\frac{1}{16}\)

\(B=\frac{15}{16}\)

Vậy \(B=\frac{15}{16}\)

Chúc bạn học tốt ~

Ta có : 

\(C=\frac{2}{1.2}+\frac{4}{2.4}+\frac{6}{4.7}+\frac{8}{7.11}+\frac{10}{11.16}+\frac{12}{16.22}\)

\(C=2\left(\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}+\frac{6}{16.22}\right)\)

\(C=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{22}\right)\)

\(C=2\left(1-\frac{1}{22}\right)\)

\(C=2-\frac{1}{11}\)

\(C=\frac{21}{11}\)

Vậy \(C=\frac{21}{11}\)

Chúc bạn học tốt ~ 

Câu 1:

A)

a) Để \(\frac{-5}{n-2}\)đạt giá trị nguyên thì \(-5⋮n-2\)

Vì \(-5⋮n-2\Rightarrow n-2\inƯ\left(-5\right)=\left(\pm1;\pm5\right)\) 

Ta có bảng giá trị:

Đối chiếu điều kiện \(n\inℤ\Rightarrow n\in\left(3;7;1;-3\right)\)

Đến câu b,c cậu cũng lí luận để chứng minh tử phải chia hết cho mẫu, còn tớ chỉ cần tách và đưa ra kết quả thôi nhé

b) Ta có:                   \(n-5⋮n+1\)

\(\Rightarrow\left(n+1\right)-6⋮n+1\)

\(\Rightarrow-6⋮n+1\)

Vì \(-6⋮n+1\Rightarrow n+1\inƯ\left(-6\right)=\left(\pm1;\pm2;\pm3;\pm6\right)\)

Ta có bảng giá trị:

Đối chiếu điều kiện \(n\inℤ\Rightarrow\left(0;1;2;5;-2;-3;-4;-7\right)\)

c) Ta có:                      \(3n+7⋮n-1\)

\(\Rightarrow3\left(n-1\right)+10⋮n-1\)

\(\Rightarrow10⋮n-1\)

Vì \(10⋮n-1\Rightarrow n-1\inƯ\left(10\right)=\left(1;-1;2;-2;5;-5;10;-10\right)\)

Ta có bảng giá trị:

Đối chiếu điều kiện \(n\inℤ\Rightarrow n\in\left(2;0;3;-1;6;-4;11;-9\right)\)

B)

a) Gọi d là ƯC (2n+1;2n+2) \(\left(d\inℤ\right)\)

\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\2n+2⋮d\end{cases}}\)    \(\Rightarrow\left(2n+2\right)-\left(2n+1\right)⋮d\)     \(\Rightarrow1⋮d\)

                                                                                                            \(\Rightarrow d=1\)

\(\Rightarrow\)2n+1 và 2n+2 nguyên tố cùng nhau

\(\Rightarrow\frac{2n+1}{2n+2}\)là phân số tối giản

b) Gọi d là ƯC(2n+3;2n+5) \(\left(d\inℤ\right)\)

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+5⋮d\end{cases}}\)        \(\Rightarrow\left(2n+5\right)-\left(2n+3\right)⋮d\) \(\Rightarrow2⋮d\) \(\Rightarrow d=\left(1;2\right)\)

Vì 2n+3 và 2n+5 không chia hết cho 2

\(\Rightarrow d=1\)

\(\Rightarrow\)2n+5 và 2n+3 nguyên tố cùng nhau

\(\Rightarrow\frac{2n+3}{2n+5}\)là phân số tối giản

Ta có: B = \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}\)

=> B =  \(\frac{6}{3.5}\)+ \(\frac{6}{5.7}\)+ \(\frac{6}{7.9}\)+ \(\frac{6}{9.11}\)

=>B =\(3.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

=> B = \(3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

=> B = \(3.\left(\frac{1}{3}-\frac{1}{11}\right)\)

=> B = \(3.\frac{8}{33}\)

=> B = \(\frac{8}{11}\)

Vậy: B = \(\frac{8}{11}\)

fgghhh

f

D= 1/1 - 1 /2 + 1/2 - 1/4 + 1/4 - 1/7 +...+ 1/46 - 1/56

D= 1/1 - 1/56

D=  55/56

vậy D= 55/56

Đặt \(A=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\)

\(1A=1-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}\right)-\dfrac{1}{22}\)\(1A=1-\dfrac{1}{22}\)

\(1A=\dfrac{22}{22}-\dfrac{1}{22}\)

\(1A=\dfrac{21}{22}\)

\(\dfrac{21}{22}\) không thể rút gọn

\(A=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{6}{16\cdot22}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}\\ =1-\dfrac{1}{22}\\ =\dfrac{21}{22}\)

Vậy \(A=\dfrac{21}{22}\)

P=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{46}-\frac{1}{56}\)

P=\(1-\frac{1}{56}\)

P=\(\frac{55}{56}\)

P = 1 - 1/2 + 1/2 - 1/4 +.......+1/46 - 1/56

P = 1 - 1/56

P = 55/56 nha!

dễ thôi 

P= 1/1.2 + 2/2.4 + 3/4.7 + ... + 10/46.56 

P= 1- 1/2 + 1/2 - 1/4 + 1/4 -1/7 +... + 1/46 - 1/56

P=1 - 1/56

P=55/56

x = \(\frac{163}{528}\)

cho 1 đ-ú-n-g nha bạn