+ 2 vào cả hai vế đi bạn rồi quy đồng từng phân số với 1

trên tử sẽ xuất hiện x + 10

chuyển hết sang một bên sẽ xuất hiện (x+10)*(...) = 0

do cái (...) khác 0 nên x + 10 = 0

=> x = -10

\(\dfrac{x+3}{7}+\dfrac{x+5}{5}=\dfrac{x-1}{11}+\dfrac{x-3}{13}\)

=>\(\left(\dfrac{x+3}{7}+1\right)+\left(\dfrac{x+5}{5}+1\right)=\left(\dfrac{x-1}{11}+1\right)+\left(\dfrac{x-3}{13}+1\right)\)

=>\(\dfrac{x+10}{7}+\dfrac{x+10}{5}=\dfrac{x+10}{11}+\dfrac{x+10}{13}\)

=>\(\left(x+10\right)\left(\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{11}-\dfrac{1}{13}\right)=0\)

=>x+10=0

=>x=-10

<=> 2x(25x² - 1) = 0

TH1: x = 0

TH2: 25x²-1 = 0

<=> 25x² = 1

<=> x = 1/5 hoặc -1/5

Vậy x = 0 hoặc x = 1/5 hoặc x = -1/5

=x³x(50-2)=0

=x³x48=0

=x³=0

=x=0

Vậy x =0

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

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^