Gọi độ dài các cạnh của tam giác đó lần lượt là a;b;c \(\left(a;b;c>0\right)\)

Theo gt ta có:

\(\left\{{}\begin{matrix}\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{7}\\a+b=32\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{7}=\dfrac{a+b}{3+5}=\dfrac{32}{8}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a=4.3=12\left(cm\right)\\b=4.5=20\left(cm\right)\\c=4.7=28\left(cm\right)\end{matrix}\right.\)

Vậy.........................

Ta có:

2=2

3=3

4=2²

5=5

6=2.3

=> BCNN (2;3;4;5;6) = 2².3.5 = 60

=> BC (2;3;4;5;6) = B (60) = {0;60;120;180;240;300;...}

Tick nha

Ta có:

\(B=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)

\(\Rightarrow\dfrac{1}{2}B=\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-....+\dfrac{1}{2^{100}}-\dfrac{1}{2^{101}}\)

\(\Rightarrow B+\dfrac{1}{2}B=\left(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-...+\dfrac{1}{2^{100}}-\dfrac{1}{2^{101}}\right)\)

\(\Rightarrow\dfrac{3}{2}B=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)

\(\Rightarrow B=\dfrac{\dfrac{1}{2}-\dfrac{1}{2^{101}}}{\dfrac{3}{2}}\)