Xét tam giác ABC, có:ˆA1+ˆB+ˆC1=1800 (Tổng các góc của tam giác)Xét tam giác ADC, có:ˆA2+ˆD+ˆC2=1800 (Tổng các góc của tam giác)Xét tứ giác ABCD, có:ˆA+ˆB+ˆD+ˆC=ˆA1+ˆA2+ˆB+ˆD+ˆC1+ˆC2=(ˆA1+ˆB+ˆC1)+(ˆA2+ˆD+ˆC2)=1800+1800=3600⇒đpcm

\(3^8.9^2=3^8.\left(3^2\right)^2=3^8.3^4=3^{8+4}=3^{12}\)

Lần sau nên chọn đúng môn.

\(\dfrac{\left(0,15\right)^4}{\left(0,5\right)^5}=\left(\dfrac{0,15}{0,5}\right)^4.\dfrac{1}{0,5}=\left(\dfrac{3}{10}\right)^4.2=\dfrac{81}{10000}.2=\dfrac{81}{5000}\)

KQ = 81/5000

\(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)

        \(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) 

=>      \(\left(2x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\) 

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

TH1: 

\(2x+\dfrac{1}{2}=\dfrac{1}{2}\)

        \(2x=\dfrac{1}{2}-\dfrac{1}{2}=0\) 

          \(x=0\) 

TH2:

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

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

        \(2x=-1\) 

            \(x=\dfrac{-1}{2}\) 

      Vậy \(x\in\left\{0;\dfrac{-1}{2}\right\}\)

\(\dfrac{5}{7}:x+1=\dfrac{-15}{30}\)

\(\dfrac{5}{7}:x=\dfrac{-15}{30}-1\)

\(\dfrac{5}{7}:x=\dfrac{-45}{30}\)

\(x=\dfrac{5}{7}:\dfrac{-45}{30}\)

\(x=\dfrac{-10}{21}\)

\(\dfrac{5}{7}:x+1=-\dfrac{15}{30}\)

\(\Rightarrow\dfrac{5}{7}:x+1=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{1}{2}-1\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{5}{7}:-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{10}{21}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{9}{3}\)

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

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

\(\Rightarrow x=\dfrac{7}{2}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\\ \\ \\ \Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}=\dfrac{9}{3}=3\\ \\ \\ \Rightarrow x=3+\dfrac{1}{2}=3\dfrac{1}{2}\)

\(\left(\dfrac{9}{2}-2x\right)\cdot1\dfrac{4}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{11}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\dfrac{9}{2}-2x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}\cdot\dfrac{7}{11}=\dfrac{1}{2}\\ \\ \\ \Rightarrow2x=\dfrac{9}{2}-\dfrac{1}{2}=\dfrac{8}{2}=4\\ \\ \\ \Rightarrow x=4:2=2\)

(\(\dfrac{9}{2}\) - 2\(x\)) \(\times\) 1\(\dfrac{4}{7}\) = \(\dfrac{11}{14}\)

( \(\dfrac{9}{2}\) - 2\(x\))  \(\times\) \(\dfrac{11}{7}\) =   \(\dfrac{11}{14}\)

\(\dfrac{9}{2}\)  - 2\(x\)              = \(\dfrac{11}{14}\) : \(\dfrac{11}{7}\)

\(\dfrac{9}{2}\) - 2\(x\)               = \(\dfrac{1}{2}\)

       2\(x\)               =   \(\dfrac{9}{2}\) - \(\dfrac{1}{2}\)

       2\(x\)               = 4

         \(x\)                = 4 : 2

          \(x\)               = 2

a) Số đo \(\widehat{xAy}\) là: 90^o vì có kí hiệu vuông góc.

b) Số đo \(\widehat{x'Ay}\):

Vì \(\widehat{x'Ay}\) và \(\widehat{xAy}\) là hai góc kề bù nên

nên \(\widehat{x'Ax}\) = \(\widehat{x'Ay}\) + \(\widehat{xAy}\)

       180^o = \(\widehat{x'Ay}\) + 90^o

        \(\widehat{x'Ay}\) = 180^o - 90^o

        \(\widehat{x'Ay}\) = 90^o

c) Số đo \(\widehat{x'Ay'}\):

Vì  \(\widehat{xAy}\) và \(\widehat{x'Ay'}\) là hai góc đối đỉnh 

nên: \(\widehat{x'Ay'}\) = \(\widehat{xAy}\) = 90^o

d) Số đo \(\widehat{xAy'}\):

Vì \(\widehat{xAy'}\) và \(\widehat{x'Ay}\) là hai góc đối đỉnh

nên \(\widehat{xAy'}\) = \(\widehat{x'Ay}\) = 90^o

(2^10*3^10.2^10*3^9):(2^9*3^10)

(\(\dfrac{1}{9}\))² = (\(\dfrac{1^2}{3^2}\))²= ((\(\dfrac{1}{3}\))²)²=  (\(\dfrac{1}{3}\))⁴