a)Ta có
2014xy \(⋮\) 42
=> 201400 + xy \(⋮\) 42
=> 42.4795 +xy + 10
Do 42.4795 \(⋮\) 42
=> xy + 10 \(⋮\) 42 (1)
Mà 0 \(\le\) xy \(\le\) 99
=> 10 \(\le\) xy +10 \(\le\) 109 (2)
Từ (1) + (2) => xy + 10 = 42 hoặc xy + 10 = 82
=> xy = 32 hoặc xy = 72
b)Ta có
\(\dfrac{a}{7}\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{b+1}\)
\(\dfrac{2a}{14}\) - \(\dfrac{7}{14}\) = \(\dfrac{1}{b+1}\)
\(\dfrac{2a-7}{14}\) = \(\dfrac{1}{b+1}\)
=> (2a - 7).(b+1) = 14
Mà a; b \(\in\) Z => 2a - 7; b + 1 \(\in\) Z
=> 2a - 7; b + 1 \(\in\) Ư(14)
2a - 7 \(⋮̸\) 2
Ta có bảng
| a | 2a - 7 | b + 1 | b | So đk a;b \(\in\) Z |
| 4 | 1 | 14 | 13 | Tm |
| 7 | 7 | 2 | 1 | Tm |
| -4 | -1 | -14 | -15 | Tm |
| 0 | -7 | -2 | -3 | Tm |
Vậy cặp số (a;b) = (4;13) (7;1)
(-4;-15) (0;-3)
Giải:
a) Ta có:
\(\overline{2014xy}⋮42\)
\(\Rightarrow201400+\overline{xy}⋮42\)
\(\Rightarrow42.4795+\overline{xy}+10⋮42\)
Vì \(42.4795⋮42\Rightarrow\overline{xy}+10⋮42\) (1)
Mà \(0\le\overline{xy}\le99\)
\(\Rightarrow10\le\overline{xy}+10\le109\) (2)
Từ (1) và (2)
\(\Rightarrow\left\{\begin{matrix}\overline{xy}+10=42\\\overline{xy}+10=84\end{matrix}\right.\Rightarrow\left\{\begin{matrix}\overline{xy}=32\\\overline{xy}=74\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(3;2\right);\left(7;4\right)\)
b) Quy đồng các phân số \(\frac{a}{7};\frac{1}{2}\) ta có:
\(BCNN\left(7;2\right)=7.2=14\)
Ta có:
\(14\div7=2\)
\(14\div2=7\)
Vậy: \(\left\{\begin{matrix}\frac{a}{7}=\frac{a.2}{7.2}=\frac{2a}{14}\\\frac{1}{2}=\frac{1.7}{2.7}=\frac{7}{14}\end{matrix}\right.\)
\(\Rightarrow\frac{2a}{14}-\frac{7}{14}=\frac{2a-7}{14}=\frac{1}{b+1}\)
\(\Rightarrow b+1=14\)
\(\Rightarrow b=14-1\)
\(\Rightarrow b=13\)\((*)\)
Thay \((*)\) vào ta lại có:
\(\frac{2a-7}{14}=\frac{1}{13+1}=\frac{1}{14}\)
\(\Rightarrow2a-7=1\)
\(\Rightarrow2a=1+7\)
\(\Rightarrow2a=8\)
\(\Rightarrow a=\frac{8}{2}\)
\(\Rightarrow a=4\)
Vậy \(a=4;b=13\)
sáng mai mk có tiết toán
