K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
2
18 tháng 7 2016
a) \(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\) . Đến đấy áp dụng t/c dãy tỉ số bằng nhau : \(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{15}{7}\)
\(\Rightarrow x=\frac{15}{7}.2=\frac{30}{7}\) ; \(\Rightarrow y=\frac{15}{7}.5=\frac{75}{7}\)
b) \(\frac{x}{y}=\frac{3}{7}\Rightarrow\frac{x}{3}=\frac{y}{7}\). Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{10}{-3}\)
\(\Rightarrow x=-10\) ; \(y=-\frac{70}{3}\)
c) Sai đề vì 2x = 3y => 2x - 3y = 0 mà giả thiết lại đưa ra 2x - 3y = 15 => mâu thuẫn
d) \(\frac{x+3y}{x-2y}=\frac{2}{3}\Leftrightarrow3\left(x+3y\right)=2\left(x-2y\right)\)
\(\Leftrightarrow3x+9y=2x-4y\Leftrightarrow x=-13y\)
Thay x = -13y vào x+2y = 1 được :
x + 2y = 1 => (-13y) + 2y = 1 => -11y = 1 => y = -1/11
=> x = -1/11 . -13 = 13/11
18 tháng 7 2016
Câu b) mình có nhầm xíu : \(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{10}{-4}=-\frac{5}{2}\)
\(\Rightarrow x=-\frac{15}{2};y=-\frac{35}{2}\)
TT
0
5 tháng 2 2020
a.
xy + 3x - 2y - 6 = 5
=>x(y + 3) - 2(y + 3) = 5
=>(x - 2)(y + 3) = 5.
Vì x, y thuộc Z nên x - 2, y + 3 thuộc Z
=> x - 2, y + 3 thuộc ước nguyên của 5
Lập bảng :
| x - 2 | -5 | -1 | 1 | 5 |
| y + 3 | -1 | -5 | 5 | 1 |
| x | -3 | 1 | 3 | 7 |
| y | -4 | -8 | 2 | -2 |
Vậy ......
b. Làm tương tự câu a.
c. Ta có x + y = 3 và x - y = 15
Bài này là tổng hiệu của cấp 1, áp dụng cách làm đó thì ta được số lớn là x = (3 + 15) : 2 = 9
Số bé là y = 9 - 15 = -6
d. Ta có : |x| + |y| = 1
=>|x| = 1 - |y|
Vì |x|, |y| >= 0 và |x| = 1 - |y| nên 0 =< |x|, |y| =< 1
Vì x, y thuộc Z nên x = 0 thì y = 1 hoặc -1 và ngược lại y = 0 thì x = 1 hoặc -1
14 tháng 1
nguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
N
0
12 tháng 3 2020
\(a,5x+12⋮x+1\)
\(5\left(x+1\right)+7⋮x+7\)
\(7⋮x+7\)
\(\Rightarrow x+7\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bn tự lập bảng nha , b;c ko phải mk lười , bn lm mới cs ý nghĩa , cố gắng nha !
12 tháng 3 2020
a) ta có 5x+12=5(x+1)+7
=> 7 chia hết cho x+1
x thuộc Z => x+1 thuộc Z => x+ 1 thuộc Ư (7)={-7;-1;1;7}
Ta có bảng
| x+1 | -7 | -1 | 1 | 7 |
| x | -8 | -2 | 0 | 6 |
b) (x-1) (2y-3)=4
x,y thuộc Z => x-1; 2y-3 thuộc Z => x-1; 2y-3 thuộc Ư (4)={-4;-2;-1;1;2;4}
Ta có bảng
| x-1 | -4 | -2 | -1 | 1 | 2 | 4 |
| x | -3 | -1 | 0 | 2 | 3 | 5 |
| 2y-3 | -1 | -2 | -4 | 4 | 2 | 1 |
| y | 1 | \(\frac{1}{2}\) | \(\frac{-1}{2}\) | \(\frac{7}{2}\) | \(\frac{5}{2}\) | 2 |
TH
1
NV
0
ta có \(5x-2y=87\Rightarrow x=\frac{87+2y}{5}=\frac{85+2+2y}{5}=17+\frac{2\left(y+1\right)}{5}\)(1)
để \(x\in z\Leftrightarrow2\left(y+1\right)⋮5\Rightarrow y+1⋮5\)
\(\Leftrightarrow y+1\in B\left(5\right)\)nên có dạng \(y+1=5k\)\(\left(k\in z\right)\)\(\Rightarrow y=5k-1\)
thay y=5k-1 vào (1) \(\Rightarrow x=17+2t\)
vậy tập nghiêm x, y có dạng y=5k-1 và x=17+2t