):< no

thanks

ok 

mơ đê :>>

Ta có: \(2+\frac{3}{4-\frac{1}{2-y}}=7\)

=>\(\frac{3}{4-\frac{1}{2-y}}=7-2=5\)

=>\(4-\frac{1}{2-y}=\frac35\)

=>\(\frac{1}{2-y}=4-\frac35=\frac{17}{5}\)

=>\(2-y=\frac{5}{17}\)

=>\(y=2-\frac{5}{17}=\frac{29}{17}\)

ảnh mô

OK

:'')

B1 dùng đối đỉnh c2 dùng tổng các góc bằng 360° gì đấy

a)

  + Ta có: \(\widehat{B_3}\)= 80^o ( Theo đề bài)

     ⇒\(\widehat{B_1}=\widehat{B_3}=80^o\) (Đối đỉnh)

 +Vì \(\widehat{B_3}\) và \(\widehat{C_2}\) ở vị trí trong cùng phía nên ta có:

               \(\widehat{B_3}+\widehat{C_2}=180^o\)

        Hay 80\(^o\)+\(\widehat{C_2}\)= 180\(^o\)

          ⇒          \(\widehat{C_2}=180^o-80^o\)

          ⇒           \(\widehat{C_2}=100^o\)

b)

 +Ta có: \(\widehat{C_2}+\widehat{BCy'}=180^o\) (Kề bù)

   Hay     \(100^o+\widehat{BCy'}=180^o\)

    ⇒                   \(\widehat{BCy'}=180^o-100^o\)

                          \(\widehat{BCy'}=80^o\)

 +Vì Ct là tia phân giác của \(\widehat{BCy'}\) nên ta có:

    \(\widehat{BCE}=\widehat{ECy'}=\dfrac{1}{2}\widehat{BCy'=\dfrac{1}{2}.}\)\(80^o\)= \(40^o\)

+Qua E kẻ tia m, m // CB

 Ta có: \(\widehat{ABC}=\widehat{BEm}=80^o\) (Đồng vị)

⇒\(\widehat{BEC}=80^o-40^o=40^o\)

⇒\(\widehat{BCE}=\widehat{BEC}\)

\(\dfrac{\dfrac{3}{5}.7^2-3.5^6+\dfrac{3}{5}.3^9}{\dfrac{3}{4}.7^2-\dfrac{3}{4}.5^7+\dfrac{3}{4}.3^9}=\dfrac{\dfrac{3}{5}.7^2-\dfrac{3}{5}.5^7+\dfrac{3}{5}.3^9}{\dfrac{3}{4}.7^2-\dfrac{3}{4}.5^7+\dfrac{3}{4}.3^9}\)

\(=\dfrac{\dfrac{3}{5}\left(7^2-5^7+3^9\right)}{\dfrac{3}{4}\left(7^2-5^7+3^9\right)}=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{3}{5}.\dfrac{4}{3}=\dfrac{4}{5}\)

\(=\dfrac{3\left(\dfrac{1}{5}\cdot49-5^6+\dfrac{1}{5}\cdot3^9\right)}{3\left(\dfrac{1}{4}\cdot7^2-\dfrac{1}{4}\cdot5^7+\dfrac{1}{4}\cdot3^9\right)}=\dfrac{3\cdot\dfrac{-58393}{5}}{3\cdot\dfrac{-58393}{4}}\)

\(=\dfrac{4}{5}\)

a: Ta có: \(\hat{x^{\prime}Mz}=\hat{y^{\prime}NM}\left(=120^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên xx'//yy'

b: Qua B, kẻ tia BE nằm giữa hai tia BA và BC sao cho BE//xx'//y'y

BE//AM

=>\(\hat{ABE}=\hat{xAB}\) (hai góc so le trong)

=>\(\hat{ABE}=35^0\)

BE//CN

=>\(\hat{EBC}+\hat{BCN}=180^0\) (hai góc trong cùng phía)

=>\(\hat{EBC}=180^0-125^0=55^0\)

Ta có: tia BE nằm giữa hai tia BA và BC

=>\(\hat{ABC}=\hat{ABE}+\hat{CBE}=35^0+55^0=90^0\)

=>BA⊥BC

\(a,\Leftrightarrow3^{x-2}\left(18+3^2\right)=243\\ \Leftrightarrow3^{x-2}=243:27=9=3^2\\ \Leftrightarrow x-2=2\Leftrightarrow x=4\)

\(18\cdot3^{x-2}+3^x=243\)

\(\Leftrightarrow3^{x+1}=243\)

hay x=4