Kẻ Cp//Bm

\(\Rightarrow\widehat{BCp}=180^0-\widehat{CBm}=30^0\) (trong cùng phía)

\(\Rightarrow\widehat{DCp}=50^0-30^0=20^0\\ \Rightarrow\widehat{DCp}+\widehat{CDn}=180^0\)

Mà 2 góc này ở vị trí TCP nên Cp//Dn

Vậy Bm//Dn

Kẻ Cz//Bm ta có: \(\widehat{mBC}+\widehat{BCz}=180^o\Rightarrow\widehat{BCz}=30^o\)

\(Tacó:\widehat{BCD}=\widehat{BCz}+\widehat{zCD}\Rightarrow\widehat{zCD}=20^o\)

\(\widehat{zCD}+\widehat{CDn}=20^o+160^o=180^o\)

Mà 2 góc này là 2 góc trong cùng phía ⇒Cz//Dn

Cz//Bm, Cz//Dn⇒BM//DN

xấu đều còn hơn tốt lỏi

xấu đều còn hơn tốt lỏi

~thi tốt nhé~

\(\widehat{B_2}=\widehat{B_4}=60^0\left(đối.đỉnh\right)\\ \widehat{B_2}+\widehat{B_1}=180^0\left(kề.bù\right)\\ \Rightarrow\widehat{B_1}=180^0-60^0=120^0\\ \Rightarrow\widehat{B_3}=\widehat{B_1}=120^0\left(đối.đỉnh\right)\)

Vì a//b nên \(\widehat{B_2}=\widehat{A_4}=60^0;\widehat{B_1}=\widehat{A_3}=120^0\left(so.le.trong\right)\)

Ta có \(\left\{{}\begin{matrix}\widehat{A_2}=\widehat{A_4}=60^0\\\widehat{A_1}=\widehat{A_3}=120^0\end{matrix}\right.\left(đối.đỉnh\right)\)

1h20'=1&1/3h

\(2n+3⋮3n+4\Leftrightarrow6n+9⋮3n+4\)

\(\Leftrightarrow2\left(3n+4\right)+1⋮3n+4\Leftrightarrow1⋮3n+4\)

\(\Rightarrow3n+4\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(2n+3⋮3n+4\)

Ta có: \(2n+3=3\left(2n+3\right)=6n+9\)

\(3n+4⋮3n+4\Leftrightarrow2\left(3n+4\right)⋮3n+4\Leftrightarrow6n+8⋮3n+4\Leftrightarrow\left(6n+9\right)-\left(6n+8\right)⋮3n+4\)

\(\Leftrightarrow1⋮3n+4\Leftrightarrow3n+4\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow n\in\left\{-1;\frac{-5}{3}\right\}\)

( - 25 ) : 10 = -15 : x

=> -15 : x = - 25 : 10

     -15 : x = - 2,5

     x          = -15 : -2,5

     x          = 6

Vậy x = 6

Ta có: (-25) : 10 = (-15) : x

=>      -2,5          = (-15) : x

=>       x             = (-15) : (-2,5)

=>       x             = 6

Vậy x = 6 

-25 : 10 = -15 : x

=> -25x = -15 . 10

-25x = -150

x= 6

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