\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{300}}\\ 3S=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{300}}\right)\\ 3S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{299}}\\ 3S-S=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{299}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{300}}\right)\\ 2S=1-\dfrac{1}{3^{300}}\\ S=\dfrac{1-\dfrac{1}{3^{300}}}{2}\)

Vậy \(S=\dfrac{1-\dfrac{1}{3^{300}}}{2}\)

`3/2 -x = 4/9`

`x = 3/2 - 4/9`

`x= 19/18`.

\(k)\left(-2\right)^3\cdot\dfrac{-1}{24}+\left(\dfrac{4}{5}-1,2\right):\dfrac{2}{15}\\ =-8\cdot\dfrac{-1}{24}+\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\cdot\dfrac{15}{2}\\ =\dfrac{\left(-8\right)\cdot\left(-1\right)}{24}+\dfrac{-2}{5}\cdot\dfrac{15}{2}\\ =\dfrac{1}{3}+\left(-3\right)\\ =\dfrac{1}{3}-\dfrac{9}{3}\\ =-\dfrac{8}{3}\)

\(l)25\%-1\dfrac{1}{2}-\left(-\dfrac{1}{2}\right)^2+0,25:\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{3}{2}-\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{12}{1}\\ =\dfrac{1}{4}-\dfrac{6}{4}-\dfrac{1}{4}+3\\ =-\dfrac{3}{2}+\dfrac{6}{2}\\ =\dfrac{3}{2}\)

\(m)\left(\dfrac{-2}{5}\right)^2+\dfrac{1}{2}\cdot\left(4,5-2\right)-50\%\\ =\dfrac{4}{25}+\dfrac{1}{2}\cdot\left(\dfrac{9}{2}-\dfrac{4}{2}\right)-\dfrac{1}{2}\\ =\dfrac{4}{25}+\dfrac{5}{4}-\dfrac{1}{2}\\ =\dfrac{16}{100}+\dfrac{125}{100}-\dfrac{50}{100}\\ =\dfrac{91}{100}.\)

\(\dfrac{1}{2}\) x \(\dfrac{4}{3}\) x 10 x \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\) = \(\dfrac{1}{2}\) x 2 = 1

(rút gọn 4/3 và 3/4 rồi rút 1/5 với 10, cuối cùng rút 1/2 và 2)

\(\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{10\cdot10}{9\cdot11}\)

\(=\dfrac{2\cdot3\cdot...\cdot10}{1\cdot2\cdot...\cdot9}\cdot\dfrac{2\cdot3\cdot...\cdot10}{3\cdot4\cdot...\cdot11}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{11}=\dfrac{20}{11}\)

\(\dfrac{2^{10}\cdot3^8-6^8}{4^4\cdot9^5}\)

\(=\dfrac{2^{10}\cdot3^8-2^8\cdot3^8}{2^8\cdot3^{10}}\)

\(=\dfrac{2^8\cdot3^8\left(2^2-1\right)}{2^8\cdot3^{10}}=\dfrac{1}{3^2}\cdot3=\dfrac{1}{3}\)

a) AI là trung điểm của đoạn AB => AI = 1/2 AB
=> AB = 8 x 2 = 16cm
b) Góc vuông: tMz^; tMA^
Góc tù: yAM^; xAz^
Góc nhọn: yAx^
Góc bẹt: yHA^; AMz^

 M = \(\dfrac{18-4n}{n-3}\) (n \(\in\) Z)

M \(\in\) Z ⇔ 18 - 4n ⋮ n - 3

                6 - (4n - 12) ⋮ n - 3

                6 - 4.(n - 3) ⋮ n - 3

                6 ⋮ n - 3

           n - 3 \(\in\) Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

           n  \(\in\) {-3; 0; 1; 2; 4; 5; 6; 9}

Vậy để M = \(\dfrac{18-4n}{n-3}\) có giá trị nguyên thì n \(\in\){-3; 0; 1; 2; 4; 5; 6; 9}

a: \(B=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{1}{6}\right)+...+\left(-\dfrac{1}{90}\right)\)

\(=-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)

b: \(D=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\cdot4}\)

=>\(\dfrac{D}{7}=\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\)

=>\(\dfrac{D}{7}=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{15}-\dfrac{1}{28}=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{13}{28}\)

=>\(D=\dfrac{13}{4}\)

Lời giải:

$\frac{x-2}{5}=\frac{-2}{2y+1}$

$\Rightarrow (x-2)(2y+1)=5(-2)=-10$
Do $x,y$ là số nguyên nên $x-2, 2y+1$ cũng là số nguyên. Mà $2y+1$ là số lẻ nên ta xét các TH sau:

TH1: $2y+1=1, x-2=-10\Rightarrow y=0; x=-8$

TH2: $2y+1=-1, x-2=10\Rightarrow y=-1; x=12$

TH3: $2y+1=5, x-2=-2\Rightarrow y=2; x=0$

TH4: $2y+1=-5, x-2=2\Rightarrow y=-3; x=4$