Trả lời

A.12:(x+4).3=6

    12:(x+4)   =6:3

    12:(x+4)   =2

         x+4     =12:2

         x+4     =6

         x         =6-4

         x         =2

a) 12 : (x + 4).3 = 6

\(\frac{12}{\left(x+4\right)}.3=6\)

\(\frac{36}{\left(x+4\right)}=6\)

x + 4 = 36 : 6

x + 4 = 6

x = 6 - 4

x = 2

b) 12 : (x + 4).3 = 5

\(\frac{12}{\left(x+4\right)}.3=5\)

\(\frac{36}{\left(x+4\right)}=5\)

x + 4 = 36 : 5

x + 4 = \(\frac{36}{5}\)

x = \(\frac{36}{5}\) - 4

x = \(\frac{16}{5}\)

c) 12 : (x + 4).3 + 18 = 24

\(\frac{12}{\left(x+4\right)}.3+18=24\)

\(\frac{18x+108}{x+4}=24\)

18x + 108 = 24x + 96

18x + 108 - 24x = 96

-6x + 108 = 96

-6x = 96 - 108

-6x = -12

x = -12 : -6

x = 2

a) 12 : (x + 4) . 3 = 6

=>(x + 4) . 3 = 12 : 6

=> (x + 4) . 3 = 2

=> x + 4         = 2 . 3

=> x + 4         = 6

=> x               = 2

b) 12 : (x + 4) . 2 = 5 

    12 : (x + 4)      = 5 : 2

     12 : (x + 4)     = 2,5

             x + 4      = 12 : 2,5

             x + 4      = 4,8

             x             = 0,8

c) 12 : (x + 4) . 3 + 18 = 24

=>12 : (x + 4) . 3          = 42

=> 12 : (x + 4)              = 42 : 3

=> 12 : (x + 4)              = 14

=>         x + 4                = 12 : 14

=>         x + 4                 = 6/7

=>          x                       = 6/7 - 4

=>          x                       = -22/7

\(\frac{8}{12}\cdot x\cdot4-\frac{18}{12}\cdot x\cdot6=\frac{5}{12}\)

\(x\left(\frac{8}{12}\cdot4-\frac{18}{12}\cdot6\right)=\frac{5}{12}\)

\(x\left(\frac{32}{12}-\frac{108}{12}\right)=\frac{5}{12}\)

\(x\left(-\frac{19}{3}\right)=\frac{5}{12}\)

\(x=\frac{5}{12}\div\left(-\frac{19}{3}\right)=-\frac{5}{76}\)

Vậy \(x=-\frac{5}{76}\)

Chúc bạn học tốt ^^!!!

v

\(\dfrac{11}{4}-\dfrac{9}{8}=\dfrac{22-9}{8}=\dfrac{13}{8}+\dfrac{15}{12}=\dfrac{156+120}{96}=\dfrac{276}{96}\)

\(\dfrac{126}{84}+\dfrac{5}{3}=\dfrac{378+420}{252}=\dfrac{798}{252}\)

\(\dfrac{13}{18}:\dfrac{5}{9}=\dfrac{13.5}{18.9}=\dfrac{615}{162}-\dfrac{1}{1}=\dfrac{615-162}{162}=\dfrac{453}{162}\)

\(\dfrac{3}{9}+\dfrac{45}{12}=\dfrac{36+405}{108}=\dfrac{441}{108}\)

a,11/4−9/8+15/12

=    13/8    + 15/12 

=  276/96

= 29/8

b,6/7x21/12+5/3

=  126/84 +  5/3

=   798/252

=         

c,13/18:5/9−1

=   615/162- 1

=  453/162

d,3+9/4:3/5

=  3 + 45/12

= 441/108

a) ( x - 4 ) . ( x - 3 ) = 0

Thỏa mãn điều kiện x - 4 hoặc x - 3 bằng 0 .

\(\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}\)

b) ( x - 4 ) . ( y - 3 ) = 0

Thỏa mãn điều kiện x - 4 hoặc y - 3 bằng 0 .

\(\Rightarrow\hept{\begin{cases}x=4,x=3\\y=4,y=3\end{cases}}\)

c) x - 24 : 12 = 15

x - 2 = 15

x = 13

d) ( x - 12 ) : 13 = 5

x - 12 = 65

x = 77

e) 1 + 3 + 5 + ..... + x = 36

..... + x = 27

..... + x = 20

x = 11

f) 2 + 4 + .... + 2x = 110

2 + 4 + ....x = 55

Thỏa mãn điều kiện \(x\in N\left|2x+0\right|x⋮2\)

\(x\in\varnothing\)

\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)

\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)

\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)

\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)

12,7:4 = 3,175 ; 12:6,13 = 1,957585644 ; 9:3,5*4,15 = 10,67142857

12/17 : 4 = 12/17 * 1/4 = 12/68 = 6/34

12 : 6/13 = 12/1 * 13/6 = 156/6 = 26

9 : 3/5 * 4/15 = 9/1 * 5/3 * 4/15 = 15/1 * 4/15 = 60/15 = 4

Câu 1 là : 12/17:4=12/17:4/1=12/17*1/4=12/68=3/17

Câu 2 là : 12:6/13=12/1:6/13=12/1*13/6=156/6=26

Câu 3 là : 9:3/5*4/15=9/1:3/5=9/1*5/3=27/25*4/15=36/25

trl giùm mk đi =(((

\(1.\) \(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(=>-12x+60+21-7x=5\)
\(=>-12x+81-7x=5\)
\(=>-12x-7x+81=5\)
\(=>-19x+81=5\)
\(=>-19x=-76\)
\(=>x=4\)
\(2.\) \(\left(x-2\right).\left(x+15\right)=0\)
\(=>\left[\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.=>\left[\begin{matrix}x=2\\x=-15\end{matrix}\right.\)
\(3.\) \(\left(7-x\right).\left(x+19\right)=0\)
\(=>\left[\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.=>\left[\begin{matrix}x=7\\x=-19\end{matrix}\right.\)
\(4.\) \(\left|x\right|< 3\)
Xét: x là số dương => x < 3
Xét: x là số âm => x < -3

1: =>-12x+60+21-7x=5

=>-19x=-76

hay x=4

2: =>x-2=0 hoặc x+15=0

=>x=2 hoặc x=-15

3: =>7-x=0 hoặc x+19=0

=>x=7 hoặc x=-19