a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

\(d,-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{13}{12}\)

\(\Leftrightarrow x=\dfrac{13}{12}\)

\(c,3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(\Leftrightarrow3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(\Leftrightarrow3x=\dfrac{8}{3}-\dfrac{3}{4}\)

\(\Leftrightarrow3x=\dfrac{23}{12}\)

\(\Leftrightarrow x=\dfrac{23}{12}:3\)

\(\Leftrightarrow x=\dfrac{23}{36}\)

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

a) 4x - 15 = -75 -x

   4x+x = -75 + 15

   5x = 60

     x= 60: 5

  => x= 12

b) 3| x-7| = 21

      |x-7|= 21:3

      |x-7|=7

  => x-7 =7 hoặc x-7=-7

 +) x-7=7

     x=7+7=14

  +) x-7=-7

      x= -7+7=0

=> x=14 hoặc x=0

c) Áp dụng t/c phân số bằng nhau 

=> x= \(\frac{-3.\left(-2\right)}{6}\)=\(\frac{6}{6}\)=1

Thay x=1 => y= \(\frac{\left(-2\right).\left(-18\right)}{1}\)=\(\frac{36}{1}\)=36

Thay y =36 => z=\(\frac{\left(-18\right).24}{36}\)=\(\frac{-432}{36}\)=-12

vậy (x,y,z)= (1;36;-12)

(câu d dài quá vs lại cx dễ nên bn tự lm nha mk chỉ giúp đến đây thôi)

Cô Hoài ơi, cô trả lời tin nhắn em với ạ!

a, (-25) + 276 - (276 - 25)

= (-25) + 276 - 276 + 25

= [ (-25) + 25] - (276 - 276)

= 0 - 0

 = 0

b, 24 x 46 + 24 x 53 + 24 

= 24 x ( 46 + 53 + 1)

= 24 x 100

= 2400

c, 145 + 20 - 45 + 180

= (145 - 45) + (20 + 180)

= 100 + 200

= 300

d; 12 x 36 + 12 x 65 - 12

= 12 x ( 36 + 65  - 1)

= 12 x (100)

= 1200

a: =>2x>-6

hay x>-3

e: =>(5-x)/x<0

=>0<x<5

h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)

\(\Leftrightarrow x+3< 0\)

hay x<-3

g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

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

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

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

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

( 2 x y + 2/15 ) x 3 = 4/5

( 2 x y + 2/15 )      = 4/5 : 3 

( 2 x y + 2/15 )      =   4/15

 2 x y                    = 4/15 - 2/15 

2 x y                     =     2/15

     y                      =     2/15 :2 

   y                          =    1/15

(2 x y + 2/15) x 3 = 4/5 

2 x y + 2/15) = 4/5 : 3 

2 x y + 2/15 = 4/15 

2 x y = 4/15 - 2/15 

2 x y = 2/15 

y = 2/15 : 2 

y = 1/15 

7/9 x (2 - 1/3 x y) = 14/15 

(2 - 1/3 x y) = 14/15 : 7/9 

(2 - 1/3 x y) = 6/5 

2 - y = 6/5 x 1/3 

2 - y = 2/5 

y = 2/5 + 2 

y = 12/5 

4/21 + 5 x y - 8/7 = 1/3 

4/21 + 5 x y = 1/3 + 8/7 

4/21 + 5 x y = 31/21 

5 x y = 31/21 - 4/21 

5 x y = 9/7 

y = 9/7 : 5 

y = 9/35 

7/12 x y - 3/12 x y = 5 

y x (7/12 - 3/12) = 5 

y x 1/3 = 5 

y = 5 : 1/3 

y = 15 

a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

 =  \(\dfrac{5}{6}\)  + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

= 1 + \(\dfrac{1}{23}\)

 = \(\dfrac{24}{23}\) 

b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)

= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5

= 1 - 1 + 0,5

= 0,5 

c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)

=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)

=0

d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)

= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)

= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)

= 1

e, 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

 =    (0,5 + 0,4) + ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + (\(\dfrac{5}{7}\)  - \(\dfrac{4}{35}\))

 = 0,9 + \(\dfrac{1}{2}\) + \(\dfrac{3}{5}\)

= 0,9 + 0,5 + 0,6

= 2

a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

Mong bạn xem lại đề bài.