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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!
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: \(\Leftrightarrow70+18< x< 120+126+70\)
=>88<x<316
hay \(x\in\left\{89;90;...;315\right\}\)
b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)
=>-3<x<3,4
hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)
1-\(\dfrac{1}{12}\)-\(\dfrac{1}{20}\)-\(\dfrac{1}{30}\)-...-\(\dfrac{1}{210}\)
=1-(\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{210}\))
=1-(\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{14.15}\))
=1-(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{14}\)-\(\dfrac{1}{15}\))
=1-(\(\dfrac{1}{3}\)-\(\dfrac{1}{15}\))
=1-\(\dfrac{4}{15}\)
=\(\dfrac{11}{15}\)
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
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