xy//BC

=>\(\widehat{xAB}=\widehat{ABC}\)(hai góc so le trong) và \(\widehat{yAC}=\widehat{ACB}\)(hai góc so le trong)

Ta có: \(\widehat{xAB}+\widehat{BAC}+\widehat{yAC}=180^0\)

=>\(\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^0\)

=>Tổng số đo 3 góc của ΔABC là 180 độ

`15/12 -(-5/13) - 3/12 - 18/13`

` = 15/12 + 5/13 - 3/12 - 18/13`

` = (15/12 - 3/12) + (5/13 - 18/13)`

`= 1 + -1`

`= 0`

\(\dfrac{x+1}{2}\) = 1 - \(x\)

 \(x+1\) = 2.(1 - \(x\))

\(x+1\) = 2 - 2\(x\)

2\(x\) + \(x\) = 2 - 1

3\(x\)        = 1

  \(x=\dfrac{1}{3}\)

Vậy \(x=\dfrac{1}{3}\)

Hình bốn đâu em?

                        Giải ta có:

         M + P + N = 180⁰ (tổng ba góc trong một tam giác bằng 180⁰

       ⇒ p + N =  180⁰ - 90⁰ = 90⁰

                 \(\dfrac{N}{P}\) = \(\dfrac{3}{2}\) ⇒ \(\dfrac{N}{3}\) = \(\dfrac{P}{2}\)

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

            \(\dfrac{N}{3}\) = \(\dfrac{P}{2}\) =  \(\dfrac{N+P}{3+2}\) = \(\dfrac{90}{5}\) = 18⁰

            N  = 18⁰ x 3  = 54⁰

            P  = 18⁰ x 2 = 36⁰

Kết luận: M = 90⁰; N = 54⁰; P =  36⁰

Em chọn vào biểu tượng \(\Sigma\) góc tái màn hình em nhé. Sau đó em nhấn biểu tượng phân số rồi em chèn phân số vào là được. 

  A = \(\dfrac{3^2}{4}\) + \(\dfrac{3^2}{18}\) + \(\dfrac{3^2}{54}\) + \(\dfrac{3^2}{108}\) + \(\dfrac{3^2}{180}\) + \(\dfrac{3^2}{270}\)

A = \(\dfrac{9}{4}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)

A = \(\dfrac{9}{4}\)  + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)

A = \(\dfrac{9}{4}\) + (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\))

A = \(\dfrac{9}{4}\) + \(\dfrac{1}{1}\) - \(\dfrac{1}{6}\)

A = \(\dfrac{54}{24}\) + \(\dfrac{24}{24}\) - \(\dfrac{4}{24}\)

A = \(\dfrac{78}{24}\) - \(\dfrac{4}{24}\)

A = \(\dfrac{37}{12}\)  

3¹⁰ hay như nào em ơi?

b: \(\left(x-\dfrac{1}{3}\right)^3=-\dfrac{8}{27}\)

=>\(\left(x-\dfrac{1}{3}\right)^3=\left(-\dfrac{2}{3}\right)^3\)

=>\(x-\dfrac{1}{3}=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}\)

c: \(\left(5x+1\right)^2=\dfrac{36}{49}\)

=>\(\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{6}{7}-1=-\dfrac{1}{7}\\5x=-\dfrac{6}{7}-1=-\dfrac{13}{7}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{1}{7}:5=-\dfrac{1}{35}\\x=-\dfrac{13}{7}:5=-\dfrac{13}{35}\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}-\dfrac{3}{2}x\right)^2=2\dfrac{1}{4}\)

=>\(\left(\dfrac{3}{2}x-\dfrac{1}{3}\right)^2=\dfrac{9}{4}\)

=>\(\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{3}=\dfrac{3}{2}\\\dfrac{3}{2}x-\dfrac{1}{3}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{3}{2}+\dfrac{1}{3}=\dfrac{11}{6}\\\dfrac{3}{2}x=-\dfrac{3}{2}+\dfrac{1}{3}=-\dfrac{7}{6}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}\cdot\dfrac{2}{3}=\dfrac{11}{9}\\x=-\dfrac{7}{6}:\dfrac{3}{2}=-\dfrac{7}{6}\cdot\dfrac{2}{3}=-\dfrac{7}{9}\end{matrix}\right.\)

e: \(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)

=>\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)

=>2x+5=4

=>2x=4-5=-1

=>\(x=-\dfrac{1}{2}\)

g: \(\left(\dfrac{1}{3}\right)^{x+1}+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{1}{12}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{9}=\dfrac{1}{12}\)

=>\(\left(\dfrac{1}{3}\right)^x\left(\dfrac{1}{3}+\dfrac{1}{9}\right)=\dfrac{1}{12}\)
=>\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{12}:\dfrac{4}{9}=\dfrac{1}{12}\cdot\dfrac{9}{4}=\dfrac{3}{4\cdot4}=\dfrac{3}{16}\)

=>\(x=log_{\dfrac{1}{3}}\left(\dfrac{3}{16}\right)\)

Mọi người giúp mk bài này với!