\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\\ \Rightarrow A=\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{20\times21}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\\\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{182}\)

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 = 1/182

1/182

Gọi tổng đó là tổng S

Ta có: S = 1/6+1/12+1/30+1/42+1/56+1/72+1/90

=>      S = 1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=>      S = 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

=>      S = 1/2-1/10

=>      S = 5/10-1/10

=>      S=4/10

=>      S=2/5 

Phân số là \(\frac{2}{5}\)

ta thấy: 1/6=1/2x3; 1/12=1/3x4;1/20=1/4x5;.............;1/90=9x10

Đặt A= 1/6+1/12+1/20+...+1/90

A=1/2x3+1/3x4+1/4x5+.....+1/9x10

A=1/2 -1/3+1/3 -1/4+1/4 -1/5+.......+1/9 -1/10

A=1/2-(1/3+1/3)-(1/4+1/4)-......-(1/9+1/9)-1/10

A=1/2-0-0-0-.....0-1/10

A=1/2-1/10

A=2/5

Sửa đề: x² + 13x + 41 --> x² + 13x + 42

Giải:

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+41}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{2}\)

(ĐKXĐ: \(x\ne\left\{-1;-2;-3;-4;-5;-6;-7\right\}\))

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{1}{2}\)

\(\Leftrightarrow\frac{x+7-x-1}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)\left(x+7\right)=12\)

\(\Leftrightarrow x^2+8x+7=12\)

⇔$x^2-8x=5$

⇔ $8x+{(-4)}^2=5+{(-4)}^2$
⇔ $x^2-8x+16=21$
⇔ ${(x-4)}^2=21$
⇔ $x=\pm \sqrt{21}+4$

Vậy...

Chúc bạn học tốt@@

vabh ơi cho mk hỏi bạn có ghi sai đề k ạ?

sr mik viết sai đề 42 ko phải 41

\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)

\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\frac{98}{99}\)

\(\Rightarrow B=\frac{98}{33}\)

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=1-\frac{1}{7}\)

\(\Rightarrow A=\frac{6}{7}\)

\(D=\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{128}\)

\(\Rightarrow2D=1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{64}\)

\(\Rightarrow2D-D=\left(1+\cdot\cdot\cdot+\frac{1}{64}\right)-\left(\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{128}\right)\)

\(\Rightarrow D=1-\frac{1}{128}\)

\(\Rightarrow D=\frac{127}{128}\)

a) \(=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(=\left[\sqrt{14}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right].\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}\)

\(=\left(\sqrt{14}+\sqrt{6}\right)\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)\)

\(=\sqrt{49}-\sqrt{21}+\sqrt{21}-\sqrt{9}\)

\(=7-3=4\)

a) \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(\left(\dfrac{14}{\sqrt{14}}+\dfrac{\sqrt{12}+\sqrt{30}}{\sqrt{2}+\sqrt{5}}\right).\sqrt{5-\sqrt{21}}=\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=4\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{199}{600}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{600}\)

=>x+1=600

hay x=599

: A = 1/6+1/12+1/20+1/30+.........+1/210

A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/14.15

A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/14 - 1/15

A = 1/2 - 1/15

A = 13/30

Ta có: 1/2= 1/1- 1/2 
1/6= 1/2 - 1/3 
1/12= 1/3- 1/4 
... 
1/30= 1/5 - 1/6 
1/42= 1/6 - 1/7 
Thay vào tổng kia: 1/2+1/6+...+1/30+1/42= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/5 - 1/6 + 1/6 - 1/7 = 1/2 - 1/7= 5/14 
Chúc bạn học tốt. Thân!

ể! Hai câu trả lời khác nhau mình nên nghe ai đây?

Nhưng mà cũng cảm ơn 2 bạn nhiều nha