xy + 3y = 66

y(x + 3 ) = 66

Vì  y( x + 3 ) = 66 nên y ; x + 3 \(\in\)Ư(66)

Ư(66) \(\in\){ 1,2,3,6,11,22,33,66}

Ta có bảng sau 

KL các cặp x;y như trên

Ta có: xy + 3y = 66

   => y.(x+3) = 66 = 1.66 = 22.3 = 33.2 = 11.6  và ngược lại ( vì x;y thuộc N)

* Nếu y = 1 thì x = 63

* Nếu y = 66 thì x = -2 (loại)

* Nếu y = 22 thì x = 0

* Nếu y = 3 thì x = 19

* Nếu y = 33 thì x = -1 (loại)

* Nếu y = 2 thì x = 30

* Nếu y = 11 thì x = 3

* Nếu y = 3 thì x = 8

Vậy ta tìm được 6 cặp x;y thuộc N

xy+3y=66

(x+3)y=66

=> y(x + 3) = 66

Mà 66 = 6 . 11 = 6(8 + 3)

=> y = 6 / x = 8

Vì xy+3y=66

=>y=3 hoặc y=8 

TH1: y=3 =>xy+3y=66

                    x3+33=66

                    x3      =66-33

                    x3      =33

=>x=3

TH2: y=8 =>xy+3y=66

                    x8+38=66

                    x8      =66-38

                    x8      =28

=>x=2

Vậy

1)

xy + x - 4y = 12

x + y(x - 4) = 12

y(x - 4) = 12 - x

\(y=\dfrac{-x+12}{x-4}\)

Vì \(x,y\inℕ\) nên

\(\left(-x+12\right)⋮\left(x-4\right)\)

\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)

\(16⋮\left(x-4\right)\)

\(\left(x-4\right)\inƯ\left(16\right)\)

\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)

\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)

\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)

\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)

2)

(2x + 3)(y - 2) = 15

\(\left(2x+3\right)\inƯ\left(15\right)\)

\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

Ta lập bảng

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)

các thầy cô ơi giúp em vs ạ mai em phải nộp r ạ!!!

x(y+2)+3y =6

=>x(y+3)+3y+9=15

=>x(y+3)+3(y+3)=15

=>(x+3)(y+3)=15

mả .....=......=>ta co bang sau

KO BIẾTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Bài 1:

a,x + ( x + 1) + (x + 2) + (x + 3) +....+ (x + 30) = 1240

x + x +x +.... + x + (1 + 2+ 3+ ....+ 30) = 1240

31x + 465 =1240

31x = 1240 - 465

31x = 775

x = 775 : 31

x = 25

b, Đề sai, bạn xem lại đề nhé.

bài 1 câu b

 1+2+3+...+x=40

\(\frac{x.\left(x+1\right)}{2}\)=40

            x.(x+1)=40.2

            x.(x+1)=80

             x.(x+1)=?

cậu viết đề sai thì phải

a, nếu x<3/2suy ra x-2<0 suy ra |x-2|=-(x-2)=2-x

                            (3-2x)>0 suy ra|3-2x|=3-2x

ta có: 2-x+3-2x=2x+1 

        5-3x=2x+1

        5-1=2x+3x

        6=6x nsuy ra x=6(loại vì ko thuộc khả năng xét)

nếu \(\frac{3}{2}\le x<2\)thì x-2<0 suy ra|x-2|=-(x-2)=2-x

                                2-2x<0 suy ra|3-2x|=-(3-2x)=2x-3

ta có:2-x+2x-3=2x+1

      -1+x=2x+1

      -1-1=2x-x

       -2=x(loại vì ko thuộc khả năng xét)

nếu \(x\ge2\)thì x-2\(\ge\)0suy ra:|x-2|=x-2

                       3-2x<0 suy ra:|3-2x|=-(3-2x)=2x-3

ta có:x-2+2x-3=2x+1

        3x-5=2x+1

       3x-2x=5+1

     x=6(chọn vì thuộc khả năng xét)

suy ra x=6

c)\(tacó:2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\)  

   \(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{10}=\frac{z}{8}\)

suy ra:\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=k\Rightarrow x=15k;y=10k;z=8k\)

 ta có: 4(15k)-3(10k)+5(8k)=7

           60k-30k+40k=7

           70k=7 suy ra k=1/10

ta có:x=1/10.15=3/2

        y=1/10.10=1

tìm x biết : I x-2I + I3-2x I = 2x+1 

b) tìm x,y thuộc Z biết : xy+2x-y= 5

c) tìm x,y thuộc Z biết : 2x=3y; 4y=5z va 4x-3y + 5z = 7 

Câu hỏi tương tự Đọc thêm Báo cáo

Toán lớp 7

nhoc quay pha 22/02/2016 lúc 15:37

c)

a) x + xy + y = 9

x(y  + 1) + y = 9 

x(y + 1) + y + 1 = 9 + 1

x(y + 1) + (y + 1) = 9 + 1

(x + 1)(y + 1) = 10 = 2.5 = 1.10 = (-2)(-5) = (-1)(-10)

Liệt kê ra 

a,x+xy+y=9

<=>x+xy+y+1=10

<=>x﴾y+1﴿+﴾y+1﴿=10

<=>﴾x+1﴿﴾y+1﴿=10 =1.10=-1.(-10)=2.5=(-2).(-5)

=> +,

+,

+,

....

Từ đó ta tìm được các cặp ﴾x;y﴿thoã mãn:

﴾1;4﴿ ; ﴾0;9﴿ ; ﴾‐3;‐6﴿ ; ﴾‐2;‐11﴿ ; ﴾4;1﴿ ; ﴾9;0﴿ ; ﴾‐6;‐3﴿ ; ﴾‐11;‐2﴿

k^o biết