\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2022}\)

\(\Rightarrow\dfrac{yz+zx+xy}{xyz}=\dfrac{1}{x+y+z}\)

\(\Rightarrow\left(yz+zx+xy\right)\left(x+y+z\right)=xyz\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+3xyz-xyz=0\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Rightarrow x=-y\) hoặc \(y=-z\) hoặc \(z=-x\).

-Đến đây thôi bạn, câu hỏi sai rồi ạ.

`1/(1 xx2) + 1/(2 xx3) + ...+ 1/(9 xx 10) xx x = 27/10`

`(1- 1/2 + 1/2-1/3+...+ 1/9-1/10)  xx x = 27/10`

`(1-1/10) xx x = 27/10`

`9/10 xx x = 27/10`

`x= 27/10 : 9/10`

`x = 27/10 . 10/9`

`x= (27.10)/(10.9)`

`x=27/9 = 3`

Đáp số: x = 3

Sửa đề: \(x+\frac{1}{2021\cdot2022}+\frac{1}{2020\cdot2021}+\cdots+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}=1\)

=>\(x+1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2021}-\frac{1}{2022}=1\)

=>\(x+1-\frac{1}{2022}=1\)

=>\(x-\frac{1}{2022}=0\)

=>\(x=\frac{1}{2022}\)

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Sửa đề:

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}=\frac{9}{10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(1-\frac{1}{x}=\frac{9}{10}\)

\(\frac{1}{x}=1-\frac{9}{10}=\frac{1}{10}\)

Vậy, x = 10.

Ko bt có right ko?

Nhầm.

Chuyển \(1-\frac{1}{x}\)thành \(1-\frac{1}{x+1}\)

\(1-\frac{1}{x+1}=\frac{9}{10}\)

\(\frac{1}{x+1}=1-\frac{9}{10}=\frac{1}{10}\)

Vậy x = 10 - 1 = 9

Thế ms right chứ!

\(\frac{2021}{1\cdot2}+\frac{2021}{2\cdot3}+...+\frac{2021}{9\cdot10}=2021\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)

\(=2021\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)=2021\cdot\left(1-\frac{1}{10}\right)\)

\(=2021\cdot\frac{9}{10}=\frac{18189}{10}\)

Ta có : \(\frac{2021}{1.2}+\frac{2021}{2.3}+\frac{2021}{3.4}+...+\frac{2021}{9.10}=2021\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(=2021\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=2021\left(1-\frac{1}{10}\right)=2021.\frac{9}{10}=1818,9\)

(2/2 + 3/6 + 4/12 + 6/30) nhân 10 - x=0

(9/6+5/4) nhân 10 -x=0

11/4 nhân 10 - x= 0

2,75 nhân 10 -x=0

27.5 - x=0

x=27.5 - 0= 27.5

dùm mình nha...

1/5x6 sửa thành 1/4x5 nhé!

(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)x10-x=0

(1-1/5)x10-x=0

4/5x10-x=0

8-x=0

x=8-0=8