\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

⇔ \(2A=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)\)

⇔ 2A = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)

⇔ 2A = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

⇔ 2A = \(\dfrac{1}{3}-\dfrac{1}{99}\)

⇔ 2A = \(\dfrac{32}{99}\)

⇔ A = \(\dfrac{32}{99}:2\)

⇔ A = \(\dfrac{16}{99}\)

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\(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}=\dfrac{x-2}{2019}+\dfrac{x-1}{2018}\)

⇔ \(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}-\dfrac{x-2}{2019}-\dfrac{x-1}{2018}=0\)

⇔ \(\left(1+\dfrac{x-4}{2021}\right)+\left(1+\dfrac{x-3}{2020}\right)-\left(1+\dfrac{x-2}{2019}\right)-\left(1+\dfrac{x-1}{2018}\right)=0\)⇔ \(\dfrac{x+2017}{2021}+\dfrac{x+2017}{2020}-\dfrac{x+2017}{2019}-\dfrac{x+2017}{2018}=0\)

⇔ \(\left(x+2017\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\right)=0\)

⇔ x + 2017 = 0

⇔ x = -2017

Vậy x = -2017

g) |x - 7| + 13 = 25

⇔ |x - 7| = 25 - 13

⇔ |x - 7| = 12

⇔ \(\left[{}\begin{matrix}x-7=12\\x-7=-12\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=12+7\\x=-12+7\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=19\\x=-5\end{matrix}\right.\)

a) -x + 8 = -17

⇔ -x = -17 - 8

⇔ -x = -25

⇔ x = 25

b) 35 - x = 37

⇔ x = 35 - 37

⇔ x = -2

c) -19 - x = -20

⇔ x = -19 + 20

⇔ x = 1

d) x - 45 = -17

⇔ x = -17 + 45

⇔ x = 28

e) |x + 3| = 15

⇔ \(\left[{}\begin{matrix}x+3=15\\x+3=-15\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=15-3\\x=-15-3\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=12\\x=-18\end{matrix}\right.\)

f) |x - 3| - 16 = -4

⇔ |x - 3| = -4 + 16

⇔ |x - 3| = 12

⇔ \(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=12+3\\x=-12+3\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

k) 26 - |x + 9| = -13

⇔ |x + 9| = 26 + 13

⇔ |x + 9| = 39

⇔ \(\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=39-9\\x=-39-9\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=30\\x=-48\end{matrix}\right.\)

1+2-3-4+5+6-7-8+...........+993+994-996+997+998

=1+(2-3-4+5)+(6-7-8+9)+...........+(994-996+997+998)

=1+0+0+...........+?

=>Sai đề

b,1-3+5-7+..........+97-99+101

=(-2)+(-2)+..........+(-2)+101(50 số-2)

=(-2).50+101

=-100+101=1

\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(m_{SO_2}=0,4.64=25,6\left(g\right)\)

Số phân tử SO₂ là: 0,4.6.10²³ = 2,4.10²³ = 24.10²² (phân tử)

Khí này có thể thu bằng cách đặt đứng bình vì:

d SO₂/KK = \(\dfrac{64}{29}\approx2,2>1\)

➝ SO₂ nặng hơn không khí

➝ Đẩy không khí ra ngoài, còn lại khí SO₂ ở trong bình

a;    (x-1).(y+1)=5

Nếu (x-1).(y+1)=5

suy ra:(x-1).(y+1)thuộc Ư(5){1;5;-1;-5}

Lập bảng ta được:

x-1  1     5        -1         -5

y+1  5     1        -5         -1

x       2     6        0          -4

y       4      0        -6         -2

Ta có: (x + 2)² ≥ 0 ∀ x

Dấu ''='' xảy ra khi x + 2 = 0 ⇔ x = -2

Do đó: (x + 2)² + 4 ≥ 4 > 0

⇔ \(\dfrac{3}{\left(x+2\right)^2+4}\le\dfrac{3}{4}\)

⇔ A ≤ \(\dfrac{3}{4}\)

Dấu ''='' xảy ra khi x = -2

Vậy Max A = \(\dfrac{3}{4}\) tại x = -2