a)x=N*

b)23;x=7 kmiknha

a) 39 . (x - 5 ) = 0

            x - 5 = 0 : 39

            x - 5 = 0

                x = 0 + 5 

               x = 5

b) ( 30 - x ) . 4 = 92

               30 . x = 92 : 4

               30 . x = 23

                     x = 23 : 30

                    x =\(\frac{23}{30}\)

a) 39 . ( x - 5 ) = 0

x - 5 = 0 : 39

x - 5 = 0

x = 0 + 5

x = 5

b) ( 30 - x ) . 4 = 92

30 - x = 92 : 4

30 - x = 23

x = 30 - 23

x = 7

a)9200

9200

6400

b)50

50

19,5

c)128

128

60

a)    92 : 0,01 = 9200          92 x 100 = 9200        64 : 0,01= 6400

b)    25 : 0,5 = 50           25 x 2 = 50           39 : 0,5 = 78
c)    32 : 0,25 = 128         32 x 4 = 128          15 : 0,25 = 60

a)92 : 0,01 = 9200      92 x 100 = 9200          64 : 0,01 = 6400
b)25 : 0,5 = 50            25 x 2 = 50                  39 : 0,5 = 78
c)32 : 0,25 = 128        32 x 4 = 128                 15 x 0,25 = 60

a) (x - 78) .26 = 0 => x - 78 = 0 => x = 78

b) 39 (x - 5) = 39 => x - 5 = 1 => x = 6

c) (x - 14) : 2 = 3 => x - 14 = 6 => x = 20

d) (30 - x) . 4 = 92 

=> 30 - x = 92 : 4 

=> 30 - x = 23

=> x = 30 - 23 = 7

e) 45(91 - x) = 90

=> 91 - x = 90 : 45

=> 91 - x = 2

=> x = 91 - 2 = 89

f) (2x - 6)(3x - 18) = 0

=> \(\orbr{\begin{cases}2x-6=0\\3x-18=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=6\end{cases}}\)

g) 5x + 73.21 = 73.26

=> 5x = 73.26 - 73.21

=> 5x = 73(26 - 21) = 73.5

=> x = \(\frac{73\cdot5}{5}=73\)

h) 2x - 12 - x = 0

=> 2x - x - 12 = 0

=> x - 12 = 0

=> x = 12

làm rõ giùmnha

làm rõ giùm mình với

a) \(\left(x-78\right).26=0\)                                                    b)  \(39.\left(x-5\right)=39\)

\(\left(x-78\right)=0\)                                                                  \(x-5=39:39\)

\(x=0+78\)                                                                        \(x=1+5\)

\(x=78\)                                                                                 \(x=6\)

P/S: Dấu "." là dấu nhân nhé

\(\left(x-14\right):2=3\)

\(x-14=2.3\)

\(x=6+14\)

\(x=20\)

\(\left(30-x\right):4=92\)

\(30-x=92.4\)

\(30-x=368\)

\(x=30-368\)

\(x=-338\)

\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

=> x =  - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

nên x+100=0

hay x=-100

Vậy: S={-100}

Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)

<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)

<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)

<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)

Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)

Nên x+100=0 => x=-100 

8 ⋮ \(x\);  12 ⋮  \(x\);  24 ⋮ \(x\) ⇒ 8; 12; 24 ⋮ \(x\) ⇒ \(x\) \(\in\) ƯC(8;12;24)

8 = 2³; 12 = 2².3; 24 = 2².3

ƯCLN(8; 12; 24) =  2² = 4

\(x\) \(\in\) {-4;  -2; -1; 1; 2; 4}

a) 4
b) 6
c) 2

b, 24 ⋮ \(x\); 30 ⋮ \(x\) ⇒ \(x\) \(\in\) ƯC(24; 30) 

24 = 2³.3; 30 = 2.3.5 ⇒ ƯCLN(24; 30) = 2.3 = 6

\(x\) \(\in\) Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

a) (x - 78) . 26 = 40

=>x-78 = 40 : 26 = 20/13

=> x = 78 + 20/13

=> x = 1034/13

b) 39 . (x-5) =39

=>x-5=39:39=1

=>x=1+5=6

c) 30- ( 52 - x) =90

=>30-52+x=90

=>-22+x=90

=>x=90-(-22)

=>x=112

d) (x-12) : 5 =2

=>x-12=2.5=10

=>x=10+12=22

a, (x-78).26=40

=> (x-78)=40:26

=> x-78=40/26

=> x=40/26+78

=> x=1034/13

b, 39.(x-5)=39

=> x-5=39:39

=> x-5=1

=> x=1+5

=> x=6

c, 30-(52-x)=90

=> 52-x=30-90

=> 52-x=-60

=> x=52-(-60)

=> x=112

d, (x-12):5=2

=> x-12=2.5

=> x-12=10

=> x=10+12

=> x=22