\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)

\(\Rightarrow x+1=1991\)

\(\Rightarrow x=1990\)

Cho em cái đề bài ạ

Tìm x biết 

no help

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

Ko đề cho thêm \(\dfrac{20}{4²}\) mà 

Ta có :

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..............+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.................+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=\left(\dfrac{1}{2}+...........+\dfrac{1}{2^{10}}\right)-\left(1+\dfrac{1}{2}+..........+\dfrac{1}{2^9}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{10}}\)

\(A=\dfrac{2^{10}-1}{2^{10}}\)

~ Chúc bn học tốt ~

A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

2A=2(\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))

2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

2A-A=1-\(\dfrac{1}{2^{10}}\)=\(\dfrac{1023}{1024}\)

Đặt vế đầu là A, vế sau là B.

Vế A:

- Tử:

\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)

\(=100\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{99}+\dfrac{1}{100}\right)\)

Vậy:

\(A=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ =\dfrac{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ \Rightarrow A=50\)

Vế B:

- Tử:

\(92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\\ =\dfrac{40}{45}+\dfrac{40}{50}+...+\dfrac{40}{500}\\ =40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)\)

Vậy:

\(B=\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\\ =\dfrac{40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{500}}\\ \Rightarrow B=40\)

Từ 2 vế trên ta tính được \(\dfrac{A}{B}=\dfrac{50}{40}=\dfrac{5}{4}\)

@Tuấn Anh Phan Nguyễn giúp mk!!

Lời giải:

Xét \(1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=\frac{n^2+1}{n^2}+\frac{1}{(n+1)^2}\)

\(=\frac{(n+1)^2-2n}{n^2}+\frac{1}{(n+1)^2}=\left(\frac{n+1}{n}\right)^2+\frac{1}{(n+1)^2}-\frac{2}{n}\)

\(=\left(\frac{n+1}{n}-\frac{1}{n+1}\right)^2=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\)

\(\Rightarrow \sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n}-\frac{1}{n+1}\)

Áp dụng vào bài toán suy ra:

\(A=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)

\(=2016+\frac{1}{2}-\frac{1}{2018}=2016,5-\frac{1}{2018}\)

\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)

\(=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-3\left(\sqrt{a}-3\right)-a+2}{a-9}\)

\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\)

\(=\dfrac{11}{a-9}\)

\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)

\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)

\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)

\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(\text{đ}k\text{x}\text{đ}:a\ge0;a\ne9\right)\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a-3}\right)\left(\sqrt{a+3}\right)}-\dfrac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\dfrac{a-2}{\left(\sqrt{a}+3\right)\left(\sqrt{a-3}\right)}\\ =\dfrac{a+3\sqrt{a}-\left(3\sqrt{a}-9\right)-\left(a-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{11}{\left(\sqrt{a}-3\right)\left(\sqrt{a+3}\right)}\)

\(b,\dfrac{a+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\ =\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}\\ =\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

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

   =\(\dfrac{1}{2}:\dfrac{-3}{2}:\dfrac{4}{3}:....:\dfrac{-101}{100}\)

   =\(\dfrac{1}{2}.\dfrac{-2}{3}.\dfrac{3}{4}........\dfrac{-100}{101}\)

   =\(\dfrac{1.\left(-2\right).3......\left(-100\right)}{2.3.4...........101}\)

   =\(\dfrac{1}{101}\)

\(=\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1\)

\(=\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2008}+\dfrac{2009}{2009}\)

\(=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2009}\right)\)