K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

24 tháng 1 2020

a)x+2/11+x+2/12+x+2/13=x+2/14+x+2/15

⇒x+2/11+x+2/12+x+2/13−x+2/14−x+2/15=0

⇒(x+2)(1/11+1/12+1/13−1/14−1/15)=0

⇒x+2=0.Do 1/11+1/12+1/13−1/14−1/15≠0

⇒x=−2

19 tháng 3 2016

bạn viết sai đề bài nhé

(x+2)/11+(x+2)/12+(x+2)/13=(x+2)/14+(x+2)15 
<=> (x+2)/11+(x+2)/12+(x+2)/13 - (x+2)/14 - (x+2)/15 = 0 
<=> (x+2)(1/11+1/12+1/13 - 1/14 - 1/15 ) = 0 
vì: (1/11+1/12+1/13 - 1/14 - 1/15 ) khác 0 nên x-2 = 0 => x=2 
 

10 tháng 6 2017

Có:

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Leftrightarrow\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)

\(\Leftrightarrow\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

Dấu "=" xảy ra:

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}=0\end{matrix}\right.\)

\(\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)\ne0\)

\(\Leftrightarrow x-2=0\)

\(\Rightarrow x=0+2=2\)

Vậy \(x=2\).

Học tốt!vui

11 tháng 6 2017

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\left(\dfrac{1}{11}+\dfrac{1}{12}\right)\left(x+2\right)+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\dfrac{23\left(x+2\right)}{132}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\left(\dfrac{23}{132}+\dfrac{1}{13}\right)\left(x+2\right)=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\left(x+2\right)\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{29\left(x+2\right)}{210}\)

\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}-\dfrac{29\left(x+2\right)}{210}=0\)

\(\Rightarrow\left(\dfrac{431}{6.286}-\dfrac{29}{6.35}\right)\left(x+2\right)=0\)

\(\Rightarrow\dfrac{1}{6}\left(\dfrac{431}{286}-\dfrac{29}{35}\right)\left(x+2\right)=-2\)

25 tháng 11 2016

bạn ấy làm đúng đó bạn

25 tháng 11 2016

\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\\ \Leftrightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\\ \Rightarrow x+2=0\Leftrightarrow x=-2\)

7 tháng 3 2017

a) Ta có:

(x - 1)5 = - 243

=> (x - 1)5 = (-3)5

=> x - 1 = - 3

=> x = -3 + 1

=> x = -2

Vậy x = -2

b) Ta có:

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\Rightarrow\left(x+2\right).\dfrac{1}{11}+\left(x+2\right).\dfrac{1}{12}+\left(x+2\right).\dfrac{1}{13}=\left(x+2\right).\dfrac{1}{14}+\left(x+2\right).\dfrac{1}{15}\)

=> \(\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+2\right).\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)

=> \(\left(x+2\right).\dfrac{431}{1716}=\left(x+2\right).\dfrac{29}{210}\)

=> \(\left(x+2\right).\dfrac{431}{1716}-\left(x+2\right).\dfrac{29}{210}=0\)

=> (x + 2).(\(\dfrac{431}{1716}-\dfrac{29}{210}\)) = 0

mà \(\dfrac{431}{1716}-\dfrac{29}{210}\) \(\ne\) 0

=> x + 2 = 0

=> x = -2

Vậy x = -2

c) Ta có :

\(\left|3x-2\right|+5x=4x-10\)

=> \(\left|3x-2\right|=4x-5x-10\)

=> \(\left|3x-2\right|=-x-10\)

=> 3x - 2 = -x - 10

hoặc 3x - 2 = -(-x -10)

*) Nếu 3x - 2 = -x - 10

=> 3x + x = -10 + 2

=> 4x = -8

=> x = -2

*) Nếu 3x - 2 = -(-x -10)

=> 3x - 2 = x +10

=> 3x - x = 10 + 2

=> 2x = 12

=> x = 6

Vậy x = -2 hoặc x = 6

7 tháng 3 2017

Nguyễn Huy TúNguyễn Huy Thắngsoyeon_Tiểubàng giảiHoàng Thị Ngọc AnhAkai Haruma giúp mình bài này với

14 tháng 4 2016

BẠN VIẾT BỊ SAI ĐỀ

14 tháng 4 2016

sai ở đâu vậy bạn

22 tháng 1 2018

a, \(\left(x-1\right)^5=-243\)

\(\Leftrightarrow\left(x-1\right)^5=-3^5\)

\(\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)

b,\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)

\(\Leftrightarrow\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

\(do\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\ne0\)

\(\Rightarrow x+2=0\Leftrightarrow x=-2\)

23 tháng 1 2018

c, \(x-2\sqrt{x}=0\Leftrightarrow\sqrt{x^2}-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)