\(3.38.8+4.37.6+2.25.12\)

\(=24.38+25.37+24.25\)

\(=24\left(38+37+25\right)\)

\(=24.100=2400\)

Bài 14:

\(\widehat{BAC}=180-70-30=80^o\)(tổng 3 góc trong tam giác)

=> \(\widehat{BAC}=\widehat{ACD}\)

mà 2 góc có vị trí so le trong

=> AB//CD

Bài 15:

Kẻ AK//CD

=> \(\widehat{CAK}=180-140=40^o\)(góc trong cùng phía)

=> \(\widehat{BAC}=180^o-\widehat{CAK}=180^o-40^o=140^o\)

\(\Rightarrow\widehat{BAC}=\widehat{ACD}\)

mà 2 góc có vị trí so le trong

=> AB//CD

\(A=1+5+5^2+...+5^{2022}\)

\(5A=5+5^2+5^3+...+5^{2023}\)

\(5A-A=5+5^2+5^3+...+5^{2023}-\left(1+5+5^2+...+5^{2022}\right)\)

\(4A=5^{2023}-1\)

\(A=\dfrac{5^{2023}-1}{4}\)

\(D=2-4+6-8+...+94-96+98\)

\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+98\)

Số hạng của D là: \(\dfrac{96-2}{2}+1=48\left(số\right)\)

\(D=-2.\dfrac{48}{2}+98=50\)

Ta có:

\(A-B=C-D\)

\(\Rightarrow A+D=B+C\)

Mặt khác:

\(A+B+C+D=360^o\)

\(\Rightarrow2\left(A+D\right)=360^o\left(A+D=B+C\right)\)

\(\Rightarrow A+D=180^o\)

Mà 2 góc có vị trí trong cùng phía

=> AB//CD

=> ABCD là hình thang

\(x^2-9y^2=2022\)

\(\left(x-3y\right)\left(x+3y\right)=2022\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3y=2\\x+3y=1011\end{matrix}\right.\\\left\{{}\begin{matrix}x-3y=-2\\x+3y=-1011\end{matrix}\right.\\\left\{{}\begin{matrix}x-3y=1\\x+3y=2022\end{matrix}\right.\\\left\{{}\begin{matrix}x-3y=-1\\x+3y=-2022\end{matrix}\right.\end{matrix}\right.\)

\(\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3y=1011\\x+3y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-3y=-1011\\x+3y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x-3y=-2022\\x+3y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-3y=2022\\x+3y=1\end{matrix}\right.\end{matrix}\right.\)

Đến đây e giải 8 hệ ra là được nha

\(\left(15-6x\right).3^5=3^6\)

\(15-6x=\dfrac{3^6}{3^5}=3\)

\(6x=15-3=12\)

\(x=2\)

a) \(A=2\left(1+2+2^2+...+2^{2022}+2^{2023}\right)⋮2\left(đpcm\right)\)

b) \(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2023}\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{2023}.3\)

\(=3\left(2+2^3+...+2^{2023}\right)⋮3\left(đpcm\right)\)

Đặt \(A=2+2^2+2^3+...+2^{60}+2^{61}\)

\(2A=2^2+2^3+2^4+...+2^{61}+2^{62}\)

\(2A-A=2^2+2^3+2^4+...+2^{61}+2^{62}-\left(2+2^2+2^3+...+2^{60}+2^{61}\right)\)

\(A=2^{62}-2\)

\(=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+....+2^{92}\left(1+2+2^2+2^3\right)\)

\(=15+15.2^4+...+15.2^{92}\)

\(=15\left(1+2^4+...+2^{92}\right)⋮15\left(đpcm\right)\)