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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.
mình xin lỗi nha, câu b bài 2 đề là (x + 1)y + y mới đúng
a, 1/210+1/240+1/272+1/306
= 1/14.15 + 1/15.16 + 1/16.17 +1/17.18
= 1/14 - 1/15 + 1/15 - 1/16 +1/16 - 1/17 + 1/17 - 1/18
= 1/14 - 1/18 = 1/63
a) \(\frac{1}{210}+\frac{1}{240}+\frac{1}{272}+\frac{1}{306}\)
=\(\frac{1}{14.15}+\frac{1}{15.16}+\frac{1}{16.17}+\frac{1}{17.18}\)
=\(\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+\frac{1}{17}-\frac{1}{18}\)
=\(\frac{1}{14}-\frac{1}{18}=\frac{1}{63}\)
b)\(\frac{191}{210}+\frac{161}{240}+\frac{129}{272}+\frac{95}{306}\)
=\(\frac{191}{14.15}+\frac{161}{15.16}+\frac{129}{16.17}+\frac{95}{17.18}\)
=\(\frac{191}{14}-\frac{191}{15}+\frac{161}{15}-\frac{161}{16}+\frac{129}{16}-\frac{129}{17}+\frac{95}{17}-\frac{95}{18}\)
tự làm tiếp
giai hộ với các bạn ơi
B+4=(191/210+1)+(161/240+1)+(129/272+1)+(95/306+1)
B+4=401/210+401/240+401/272+401/306
B+4=401x(1/210+1/240+1/272+1/306)
B+4=401X 1/63
B=401/63-4
B=149/63
Hoc tot!