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22 tháng 10 2017

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 => x=2

1/ 11 + 1/12 +1/13 -1/14 - 1/15 = 0

Vì 1/11; 1/12; 1/13; 1/14; 1/15 > 1 nên 1/11+1/12+1/3-1/14-1/15= 0 (vô lí)

Vậy x=2

Nhớ like

21 tháng 10 2017

Giải:

Ta có:

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

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

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

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

Nên \(x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\).

Chúc bạn học tốt!

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

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

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

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

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

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

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

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

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

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

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

18 tháng 9 2021

Bài 1:

a) \(\left|3x-5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))

Bài 2:

a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)

b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)

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

\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)

 

18 tháng 9 2021

Bài 1:

a) \(\left|3x-5\right|=4\)  (1)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)    \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)

\(\Leftrightarrow x=-1\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)           \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

\(\Leftrightarrow x=-2004\)

25 tháng 12 2017

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

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

=> x-1= -3

=> x= -2

25 tháng 12 2017

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

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

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

=> \(\dfrac{x+2}{37}=0\)

=> x+2= 0

=> 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

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

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

`#3107`

a)

\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)

Vậy, \(x=-\dfrac{1}{5}\)

b)

\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)

Vậy, \(x=\dfrac{147}{20}\)

c)

\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)

Vậy, \(x=-\dfrac{1}{3}.\)

\(#Emyeu1aithatroi...\)

29 tháng 9 2023

(2/5 + 3/4 . x)= 11/12 -2/3

(2/5 +3/4 . x)= 1/4

3/4 . x          = 1/4 - 2/5

3/4 . x          = -3/20

x                  = -3/20 : 3/4

x                  = -1/5

Vậy .....

19 tháng 8 2017

a) Ta có : \(x - 2xy + y - 3 = 0\)

\(\Rightarrow-2xy+x+y=3\)

\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)

\(\Rightarrow4xy-2x-2y=-6\)

\(\Rightarrow4xy-2x-2y+1=-6+1\)

\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)

\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)

Tự lập bảng đi -.-

26 tháng 3 2018

Nhân từng vế bất đẳng thức ta được : (xyz)2 = 36xyz

+ Nếu một trong các số x,y,z bằng 0 thì 2 số còn lại cũng bằng 0

+ Nếu cả 3 số x,y,z khác 0 thì chia 2 vế cho xyz ta được xyz = 36

+ Từ xyz =36 và xy = z ta được z2 = 36 nên z = 6; z = -6

+ Từ xyz =36 và yz = 4x ta được 4x2 = 36 nên x = 3; x = -3

+ Từ xyz =36 và ta được 9y2 = 36 nên y = 2; y = -2

- Nếu z = 6 thì x và y cùng dấu nên x = 3, y = 2 hoặc x = -3 , y = -2

- Nếu z = -6 thì x và y trái dấu nên x = 3 ; y = -2 hoặc x = -3; y=2

Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)