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28 tháng 3 2020

\(a,x.\frac{-3}{7}=\frac{4}{21}\)

\(x=\frac{4}{21}:\frac{-3}{7}\)

\(x=\frac{-4}{9}\)

\(b,\frac{-4}{7}:x=\frac{2}{5}\)

\(x=\frac{-4}{7}:\frac{2}{5}\)

\(x=\frac{-10}{7}\)

\(c,x+\frac{1}{12}=\frac{-3}{8}\)

\(x=\frac{-3}{8}-\frac{1}{12}\)

\(x=\frac{-11}{24}\)

\(d,\frac{2}{15}-x=\frac{-3}{10}\)

\(x=\frac{2}{15}+\frac{3}{10}\)

\(x=\frac{13}{30}\)

28 tháng 3 2020

\(e,-x+\frac{4}{5}=\frac{1}{2}\)

\(-x=\frac{-3}{10}\)

\(x=\frac{3}{10}\)

\(f,\frac{3}{4}.\left(x+1\right)-\frac{1}{2}=\frac{3}{7}\)

\(\frac{3}{4}.\left(x+1\right)=\frac{13}{14}\)

\(x+1=\frac{26}{21}\)

\(x=\frac{5}{21}\)

\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)

\(\frac{-3}{2}-2x=\frac{-11}{4}\)

\(2x=\frac{-3}{2}+\frac{11}{4}\)

\(2x=\frac{-17}{4}\)

\(x=\frac{-17}{8}\)

\(h,-x+\frac{4}{5}=\frac{1}{2}\)

\(-x=\frac{-3}{10}\)

\(x=\frac{3}{10}\)

chúc bạn học tốt !!!

3 tháng 8 2021

4,  Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)

 xét x \(\ge\) \(-\frac{1}{5}\)

 Ta Có  Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\)  = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\)\(\frac{27}{35}\)   (1)

xét x \(< -\frac{1}{5}\)

Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x  + \(\frac{13}{35}\)

với x \(< -\frac{1}{5}\) 

=> -2x \(>\) \(\frac{2}{5}\) 

=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)

Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)

5 ,  D = |x| + |8-x| 

D = |x| + |8-x| \(\ge\) |x+8-x|  = |8| = 8

Dấu ''='' xảy ra khi   x(8-x) \(\ge\) 0  <=> 0\(\le\)x\(\le\) 8 

Vậy MinD = 8 khi \(0\le x\le8\) 

6,L=  |x - 2012| + |2011 - x| 

L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x |  = |-1| = 1 

Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0  

3 tháng 8 2021

làm nốt câu 6 nãy ấn nhầm 

<=> 2011\(\le\) x \(\le\) 2012

Vậy MinL = 1 khi \(2011\le x\le2012\) 

7 , E = | x- \(\frac{2006}{2007}\) | + |x-1| 

Ta có :

E = |x-\(\frac{2006}{2007}\) | + |1-x| 

E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x |  = \(\frac{1}{2007}\) 

Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=>  \(\frac{2006}{2007}\le x\le1\) 

Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\) 

8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) | 

Ta có :

F  = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\)   - x | 

F  = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x  |  = \(\frac{1}{2}\) 

Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0    <=>  \(\frac{1}{4}\le x\le\frac{3}{4}\) 

Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)

1: x=3/4-1/2=3/4-2/4=1/4

2: x-1/5=2/11

=>x=2/11+1/5=21/55

3: x-5/6=16/42-8/56

=>x-5/6=8/21-4/28=5/21

=>x=5/21+5/6=15/14

4: x/5=5/6-19/30

=>x/5=25/30-19/30=6/30=1/5

=>x=1

5: =>|x|=1/3+1/4=7/12

=>x=7/12 hoặc x=-7/12

6: x=-1/2+3/4

=>x=3/4-1/2=1/4

11: x-(-6/12)=9/48

=>x+1/2=3/16

=>x=3/16-1/2=-5/16

21 tháng 7 2023

1)x= 1/4

2)x= 2/11+ 1/5

   x= 21/55

3)x - 5/6 = 5/21

   x         = 5/21+5/6

   x         = 15/14

4)x/5 = 5/6 + -19/30

   x:5 = 1/5

   x    = 1/5.5

   x    = 1

5) |x| - 1/4 = 6/18

    |x|           = 6/18 - 1/4

    |x|            =7/12

⇒x= 7/12 hoặc -7/12

6)x = -1/2 +3/4

   x= 1/4

7) x/15 = 3/5 + -2/3

   x:15  = -1/15

  x        = -1/15. 15

  x        = -1

8)11/8 + 13/6 = 85/x  

       85/24      = 85/x

  ⇒      x           = 24

9) x - 7/8 = 13/12

   x          = 13/12 + 7/8

   x          = 47/24

10)x - -6/15 = 4/27  

     x            = 4/27 + (-6/15)

    x             = -34/135

11) -(-6/12)+x = 9/48

                    x= 9/48 - 6/12

                    x = -5/16

12) x - 4/6 = 5/25 + -7/15

      x -4/6  =  -4/15

     x           = -4/15 + 4/6

    x             = 2/5

25 tháng 7 2017

1.

a) \(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)

b) \(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)

c) \(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}=\frac{23}{2}-\frac{65}{2}=-21\)

d) \(\left(-49,1\right).\frac{13}{27}-58,9.\frac{13}{27}=\frac{13}{27}.\left(-49,1-58,9\right)=\frac{13}{27}.\left(-108\right)=-52\)

e) \(0,375:\left(-4,5\right)=\frac{-1}{12}\)

f) \(3\frac{1}{7}:\left(-1\frac{3}{7}\right)=\frac{22}{7}:\frac{-10}{7}=\frac{-11}{5}\)

g) \(9\frac{1}{3}:4\frac{2}{3}-2=\frac{28}{3}:\frac{14}{3}-2=2-2=0\)

h) \(\left(7\frac{3}{4}:0,3125+4,5.2\frac{2}{45}\right):\left(-8,5\right)=\left(\frac{31}{4}:\frac{5}{16}+\frac{9}{2}.\frac{92}{45}\right):\frac{-17}{2}=\left(\frac{124}{5}+\frac{46}{5}\right):\frac{-17}{2}=34:\frac{-17}{2}=-4\)

25 tháng 7 2017

Bài 1 : Tính:

a)

\(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)

b) 

\(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{10}{15}+\frac{-2}{15}=\frac{8}{15}\)

c)

\(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}\)\(=\frac{23}{2}-\frac{65}{2}=\frac{-42}{2}=-21\)

....

Tự lm tiếp dạng như v

Bài 2 : 

\(A=\frac{-6}{11}.\frac{7}{10}.\frac{11}{-6}.-20=\left(\frac{-6}{11}.\frac{11}{-6}\right).\left(\frac{7}{10}.-20\right)\)\(=1.\left(-14\right)=-14\)

.....

Bài 3 : 

\(\frac{3}{7}.x-\frac{2}{5}.x=\frac{-17}{35}\)

\(\Leftrightarrow\frac{3}{7}-\frac{2}{5}.x=\frac{-17}{35}\)

\(\Leftrightarrow\frac{1}{35}x=\frac{-17}{35}\)

\(\Leftrightarrow x=\frac{-17}{35}:\frac{1}{35}\)

\(\Leftrightarrow x=\frac{-17}{35}.35=-17\)

24 tháng 8 2019

a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)

vậy...

24 tháng 8 2019

2.

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

\(5x+1=\pm\frac{6}{7}\)

\(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\)\(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\)\(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)

Chúc bạn học tốt!

5 tháng 8 2017

1. So sánh

a) \(25^{50}\)\(2^{300}\)

\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)

\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)

\(25< 64\) nên \(25^{50}< 64^{50}\)

Vậy \(25^{50}< 2^{300}\)

b) \(625^{15}\)\(12^{45}\)

\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)

\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)

\(625< 1728\) nên \(625^{15}< 1728^{15}\)

Vậy \(625^{15}< 12^{45}\)

5 tháng 8 2017

1.So sánh

a)\(25^{50}\)\(2^{300}\)

Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)

\(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)

b)\(625^{15}\)\(12^{45}\)

Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)

\(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)

25 tháng 1 2020

c) \(5x-7=3x+9\)

d) \(5x-\left|9-7x\right|=3\)

e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)

h) \(5^{-1}.25^x=125\)

\(\Rightarrow\frac{1}{5}.25^x=125\)

\(\Rightarrow25^x=125:\frac{1}{5}\)

\(\Rightarrow25^x=625\)

\(\Rightarrow25^x=25^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

Chúc bạn học tốt!

25 tháng 1 2020

g) \(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)

\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{1;2;0\right\}.\)

i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)

Ta có:

\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)

\(\Rightarrow4x\ge0\)

\(\Rightarrow x\ge0.\)

Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)

\(\Rightarrow x+1+x+2+x+3=4x\)

\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)

\(\Rightarrow3x+6=4x\)

\(\Rightarrow6=4x-3x\)

\(\Rightarrow6=1x\)

\(\Rightarrow x=6\left(TM\right).\)

Vậy \(x=6.\)

Chúc bạn học tốt!

13 tháng 9 2021

 dề bài đâu mà tìm?

13 tháng 9 2021

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