60 nhe

60 nha

\(9x7x\frac{y}{42}x63=\frac{5}{6}\)
\(3969x\frac{y}{42}=\frac{5}{6}\)
\(\frac{166698}{42}x\frac{y}{42}=\frac{35}{42}\)
\(166698\)x    y=35
\(y=166698:35\)
y=4762,8

\(\frac{9\cdot7\cdot y}{72\cdot63}=\frac{5}{6}\)

Ta có : \(\frac{9\cdot7\cdot y}{9\cdot8\cdot7\cdot9}=\frac{5}{6}\)

         \(\Rightarrow\frac{1\cdot y}{8\cdot9}=\frac{5}{6}\)

           \(\Rightarrow\frac{1\cdot y}{72}=\frac{5}{6}\Rightarrow y=60\)

Do đó , y = 60

9 x 6 = 54        9 x 7 = 63        9 x 5 = 45        9 x 8 = 72

54 : 9 = 6        63 : 9 = 7        45 : 9 = 5        72 : 9 = 8

54 : 6 = 9        63 : 7 = 9        45 : 5 = 9        72 : 8 = 9

9 x 6 = 54          9 x 7 = 63          9 x 5 = 45          9 x 8 = 72

54 : 9 = 6           63 : 9 = 7           45 : 9 = 5          72 : 9 = 8

54 : 6 = 9           63 : 7 = 9           45 : 5 =9           72 : 8 =9

9 x 5 = 45

9 x 6 = 54

9 x 7 = 63

9 x 8 = 72

45 : 9 = 5

54 : 9 = 6

63 : 9 = 7

72 : 9 = 8

45 : 5 = 9

54 : 6 = 9

63 : 7 = 9

72 : 8 = 9

7 x 8 = 56       16 : 2 = 8       36 : 6 = 6       49 : 7 = 7

2 x 5 = 10       72 : 8 = 9       9 x 3 = 27       63 : 7 = 9

6 x 4 = 24       25 : 5 = 5       4 x 8 = 32       7 x 5 = 35

7 x 8 = 56

2 x 5 = 10

6 x 4 = 24

16 : 2 = 8 

72 : 8 = 9

25 : 5 = 5

36 : 6 = 6

9 x 3 = 27

4 x 8 = 32

49 : 7 = 7

63 : 7 = 9

7 x 5 = 35

a. \(\left(x+8\right)⋮\left(x+4\right)\)

\(\Rightarrow\left(x+4\right)+4⋮\left(x+4\right)\)

Mà \(\left(x+4\right)⋮\left(x+4\right)\)

\(\Rightarrow4⋮\left(x+4\right)\)

\(\Rightarrow x+4\in\text{Ư} \left(4\right)=\left\{1;2;4\right\}\)

Ta có 3 trường hợp :

TH1 : \(x+4=1\Rightarrow x\notin N\) ( Loại )

TH2 : \(x+4=2\Rightarrow x\notin N\)(Loại )

TH3 : \(x+4=4\Rightarrow x=0\)

Vậy x = 0

a,Vì : \(x+8⋮x+2\)

Mà : \(x+2⋮x+2\)

\(\Rightarrow\left(x+8\right)-\left(x+2\right)⋮x+2\Rightarrow x+8-x-2⋮x+2\)

\(\Rightarrow6⋮x+2\Rightarrow x+2\inƯ\left(6\right)\)

Mà : \(Ư\left(6\right)=\left\{1;2;3;6\right\}\) ; \(x+2\ge2\Rightarrow x+2\in\left\{2;3;6\right\}\)

\(\Rightarrow x\in\left\{0;1;4\right\}\)

Vậy ...

b,Ta có : \(2y+7⋮y-1\) ; \(y-1⋮y-1\Rightarrow2\left(y-1\right)⋮y-1\Rightarrow2y-2⋮y-1\)

\(\Rightarrow\left(2y+7\right)-\left(2y-2\right)⋮y-1\Rightarrow2y+7-2y+2⋮y-1\)

\(\Rightarrow9⋮y-1\Rightarrow y-1\in\left\{1;3;9\right\}\Rightarrow y\in\left\{2;4;10\right\}\)

Vậy ...

c, Vì : \(x\in N\Rightarrow x-5\in N\)

\(y\in N\Rightarrow y+3\in N\left(y+3\ge3\right)\)

\(\Rightarrow x-5,y+3\inƯ\left(7\right)\)

Mà : \(Ư\left(7\right)=\left\{1;7\right\};y+3\ge3\)

\(\Rightarrow x-5=1\Rightarrow x=6;y+3=7\Rightarrow y=4\)

Vậy ...

d,Vì : \(\overline{3x7y}⋮2,5\Rightarrow y=0\)

Ta có : \(\overline{3x70}⋮\) 9 dư 1

\(\Rightarrow3+x+7+0⋮\) 9 dư 1

\(\Rightarrow10+x⋮\) 9 dư 1 \(\Rightarrow9+x⋮9\)

Vì : \(9⋮9\Rightarrow x⋮9\Rightarrow x\in\left\{0;9\right\}\)

Vậy x = 0 thì b = 0

x = 9 thì b = 0

e, Vì : \(ƯCLN\left(x,y\right)=9\Rightarrow\begin{cases}x=9.k_1\\y=9.k_2\end{cases}\) với ƯCLN(k₁,k₂) = 1

Thay vào x + y = 63 ta có :

\(9.k_1+9.k_2=63\Rightarrow9\left(k_1+k_2\right)=63\Rightarrow k_1+k_2=7\)

Mà : x > y > 9

+) Nếu : k₁ = 6 ; k₂ = 1 \(\Rightarrow\begin{cases}x=6.9=54\\y=1.9=9\end{cases}\) ( loại )

+) Nếu : k₁ = 5 ; k₂ = 2 \(\Rightarrow\begin{cases}x=5.9=45\\y=2.9=18\end{cases}\) ( nhận )

+) Nếu : k₁ = 4 ; k₂ = 3 \(\Rightarrow\begin{cases}x=9.4=36\\y=9.3=27\end{cases}\) ( nhận )

Vậy x = 45 thì y = 18

x = 36 thì y = 27