1.
\(\dfrac{191}{210}+\dfrac{161}{240}+\dfrac{129}{272}+\dfrac{95}{306}=\dfrac{191}{14\cdot15}+\dfrac{161}{15\cdot16}+\dfrac{129}{16\cdot17}+\dfrac{95}{17\cdot18}=\dfrac{191}{14}-\dfrac{191}{15}+\dfrac{161}{15}-\dfrac{161}{16}+\dfrac{129}{16}-\dfrac{129}{17}+\dfrac{95}{17}-\dfrac{95}{18}=\dfrac{191}{14}-\left(\dfrac{191}{15}+\dfrac{161}{15}\right)-\left(\dfrac{161}{16}+\dfrac{129}{16}\right)-\left(\dfrac{129}{17}+\dfrac{95}{17}\right)-\dfrac{95}{18}=\dfrac{191}{14}-\dfrac{352}{15}-\dfrac{145}{8}-\dfrac{224}{17}-\dfrac{95}{18}=\dfrac{584460}{42840}-\dfrac{1005312}{42840}-\dfrac{776475}{42840}-\dfrac{564480}{42840}-\dfrac{226100}{42840}=\dfrac{-1987907}{42840}\)
2.
a) \(\left(x-5\right)\cdot y=3\)
Ta lập bảng sau:
| x-5 | 1 | 3 |
| x | 6 | 8 |
| y | 3 | 1 |
Vậy các cặp (x;y) là: (6;3) ; (8;1)
b) x+1=2
x=2-1
x=1
Vậy x=1.
Bài 2 câu 1 mk nhầm nhé bạn:
\(\left(x-5\right)\cdot y=3\)
Ta lập bảng sau:
| x-5 | 1 | 3 | -1 | -3 |
| x | 6 | 8 | 4 | 2 |
| y | 3 | 1 | -3 | -1 |
Vậy các cặp số nguyên (x;y) là: (6;3) ; (8;1) ; (4;-3) ; (2;-1).
