a) S hình thoi là:

      (19 x 12) : 2 = 114(cm2)

b) S hình thoi là;

      (30 x 7) : 2 = 105(cm2)

\(2^n.3^{2n}.\left(\frac{2}{3}\right)^n.2^n=82944\)(n\(\in\)N)

\(2^n.2^n.\left(\frac{2}{3}\right)^n.\left(3^2\right)^n=82944\)

\(\left(2.2.\frac{2}{3}.9\right)^n=82944\)

\(24^n=82944\)

Tớ làm đến đây thôi khó lắm bạn xem lại đề đi

\(2^2.3^{2n}.\left(\frac{2}{3}\right)^n.2^n=82944\)

\(2^2.9^n.\left(\frac{2}{3}\right)^n.2^n=2^{10}.3^4\)

\(2^2.2^n.\left(\frac{2}{3}.9\right)^n=2^{10}.3^4\)

\(2^{n+2}.6^n=2^{10}.3^4\)

\(2^{n+2}.2^n.3^n=2^{10}.3^4\)

\(2^{2n+2}.3^n=2^{10}.3^4\)

Vậy n = 4

\(2^2\cdot3^{2n}\cdot\left(\frac{2}{3}\right)^n\cdot2^n=82944\)

\(2^2\cdot\left(3^2\right)^n\cdot\left(\frac{2^n}{3^n}\right)\cdot2^n=82944\)

\(2^2\cdot9^n\cdot\frac{2^n}{3^n}\cdot2^n=82944\)

\(2^2\cdot\frac{9^n\cdot2^n}{3^n}\cdot2^n=82944\)

\(2^2\cdot\frac{18^n}{3^n}\cdot2^n=82944\)

\(4\cdot6^n\cdot2^n=82944\)

\(6^n\cdot2^n=82944:4\)

\(12^n=20736\)

\(12^n=12^4\)

Vậy n=4

ĐK \(n\ge0\)

Ta có \(3.3^{n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)

\(\Leftrightarrow3^n\left(6.9.3^n+3\right)-2.3^n\left(27.3^n-1\right)=405\)

\(\Leftrightarrow54.3^{2n}+3.3^n-54.3^{2n}+2.3^n=405\Leftrightarrow5.3^n=405\)

\(\Leftrightarrow3^n=81=3^4\Leftrightarrow n=4\left(tm\right)\)

Vậy \(n=4\)

\(3.3^{n-1}.\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)

\(\Rightarrow3.3^{n-1}.6.3^{n+2}+3.3.3^{n-1}-2.3^n.3^{n+3}+1.2.3^n=405\)

\(\Rightarrow3^{1+n-1}.6.3^n.3^2+3^{1+1+n-1}-2.3^n.3^n.3^3+3^n.2=405\)

\(\Rightarrow3^n.\left(6.3^2\right).3^n+3^{n+1}-\left(2.3^3\right).3^{n+n}+3^n.2=405\)

\(\Rightarrow\left(3^n.3^n\right).54+3^{n+1}-54.3^{2n}+3^n.2=405\)

\(\Rightarrow3^{2n}.54+3^{n+1}-3^{2n}.54+3^n.2=405\Rightarrow3^{n+1}+3^n.2=405\)

\(\Rightarrow3^n.3+3^n.2=405\Rightarrow3^n.5=405\Rightarrow3^n=81=3^4\Rightarrow n=4\)

xét          \(VT=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+......+\frac{1}{2n.\left(2n+2\right)}\right)\)     (1)

\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+.......+\frac{2}{2n\left(2n+2\right)}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.......+\frac{1}{2n}-\frac{1}{2n+2}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{1}{4}-\frac{1}{2\left(2n+2\right)}\)

\(=\frac{1}{4}-\frac{1}{4n+4}\)

mà theo bài ra   (1) = \(\frac{502}{2009}\)

<=>\(\frac{1}{4}-\frac{1}{4n+4}=\frac{502}{2009}\)

<=>\(\frac{1}{4n+4}=\frac{1}{4}-\frac{502}{2009}\)

<=>\(\frac{1}{4n+4}=\frac{1}{8036}\)

<=> 4n+4=8036

<=> 4n=8032

<=> n=2008

=) \(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2n\left(2n+2\right)}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}-\frac{1}{2n+2}=\frac{502}{2009}:\frac{1}{2}=\frac{1018}{2009}\)
=) \(\frac{1}{2n+2}=\frac{1}{2}-\frac{1018}{2009}=\frac{-27}{4018}\)
=) \(\frac{-1}{-\left(2n+2\right)}=\frac{-27}{4018}\)
=) \(\frac{-27}{27.-\left(2n+2\right)}=\frac{-27}{4018}\)
=) \(27.-\left(2n+2\right)=4018\)
=) \(-\left(2n+2\right)=4018:27=\frac{4018}{27}\)
=) \(2n+2=\frac{-4018}{27}\)
=) \(2n=\frac{-4018}{27}-2=\frac{-4072}{27}\)
=) \(n=\frac{-4072}{27}:2=\frac{-2036}{27}\)
\(\)
 

Ta có:

\(\frac{1}{2}.\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{2n\cdot\left(2n+2\right)}\right)=\frac{502}{2009}\)

\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)

\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2n+2}=\frac{\frac{502}{2009}}{\frac{1}{2}}=\frac{1004}{2009}\)

\(\frac{1}{2n+2}=\frac{1}{2}-\frac{1004}{2009}=\frac{1}{4018}\)

\(\Rightarrow2n+2=4018\)

\(\Rightarrow n=2013\)

Vậy n= 2013

n=1