Ta có \(\frac{y+6}{y-2}=\frac{y}{y-4}\)

=> (y + 6)(y - 4) = y(y - 2)

=> y² - 4y + 6y - 24 = y² - 2y

=> y² + 2y - 24 = y² - 2y

=> 2y - 24 = -2y

=> 2y + 2y = 24

=> 4y = 24

=> y = 6

Vậy y = 6

\(\frac{y+6}{y-2}=\frac{y}{y-4}\)

=> \(\frac{y-2+8}{y-2}=\frac{y}{y-4}\)

=> \(1+\frac{8}{y-2}=\frac{y}{y-4}\)

=> \(\frac{8}{y-2}=\frac{y}{y-4}-1=\frac{4}{y-4}\)

=> \(\frac{8}{y-2}=\frac{4}{y-4}\)

=> 8(y - 4) = 4(y - 2)

=> 8y - 32 = 4y - 8

=> 8y - 32 - 4y + 8 = 0

=> 4y - 24 = 0

=> 4y = 24

=> y = 6

Vậy y = 6

\(\frac{2x-6}{4}=\frac{x+1}{3}\)

=> \(\frac{2\left(x-3\right)}{4}=\frac{x+1}{3}\)

=> \(\frac{x-3}{2}=\frac{x+1}{3}\)

=> \(\frac{3\left(x-3\right)}{6}=\frac{2\left(x+1\right)}{6}\)

=> 3(x - 3) = 2(x + 1)

=> 3x - 9 = 2x + 2

=> 3x - 9 - 2x - 2 = 0

=> x - 11 = 0

=> x = 11

Vậy x = 11

Ta có: (2x - 6 ) : 4 = ( x + 1 ) : 3

      =>  3( 2x - 6 ) = 4 ( x + 1 )

      =>  6x - 18 = 4x + 4

      =>  2x = 22

      => x = 11

Bài 1:

Vẽ hình

Ta có: \(\widehat{xOy}+\widehat{yOt}+\widehat{zOt}+\widehat{xOz}=360^o\)(Tổng các góc trong không có điểm trong chung ) 

\(\Rightarrow\widehat{xOy}+90^o+\widehat{zOt}+90^o=360^o\)

\(\Rightarrow\widehat{xOy}+\widehat{zOt}=360^o-90^o-90^o\)

\(\Rightarrow\widehat{xOy}+\widehat{zOt}=180^o\)

Vậy \(\widehat{xOy}+\widehat{zOt}=180^o\)

Bài 2:

A) Ta có: \(\widehat{AOB}=100^o,\widehat{AOC}=90^o,\widehat{BOD}=90^o\)

\(\Rightarrow\widehat{COD}=360^o-\left(\widehat{AOB}+\widehat{AOC}+\widehat{BOD}\right)\)

\(=360^o-\left(100^o+90^o+90^o\right)=360^o-280^o=80^o\)

Ox là tia phân giác của \(\widehat{AOB}\)nên \(\widehat{xOA}=\frac{\widehat{AOB}}{2}=\frac{100^o}{2}=50^o\)

Oy là tia phân giác của \(\widehat{COD}\)nên \(\widehat{COy}=\frac{\widehat{COD}}{2}=\frac{80^o}{2}=40^o\)

Do đó: \(\widehat{xOy}=\widehat{xOA}+\widehat{AOC}+\widehat{COy}=50^o+90^o+40^o\)

Hay \(\widehat{xOy}=180^o\)

=> Ox và Oy là hai tia đối nhau ( đpcm ) 

b) Ta có: \(\widehat{xOC}=\widehat{xOA}+\widehat{AOC}=50^o+90^o=140^o\)

\(\widehat{BOy}=\widehat{BOD}+\widehat{DOy}=90^o+40^o=130^o\)

                                                         Bài giải

Ta có : \(BC\text{ }//\text{ }Az\) nên \(\widehat{C_2}=\widehat{A_2}\) ( hai góc so le trong )

Mà \(\widehat{CAx}=\widehat{A_2}+\widehat{A_3}\) là góc ngoài tại đỉnh A của \(\Delta ABC\) nên \(\widehat{A_2}+\widehat{A_3}=\widehat{B}+\widehat{C_2}\)

lại có : \(\widehat{B}=\widehat{C_2}=\widehat{A_2}\) nên \(\widehat{A_3}=\widehat{B}=\widehat{C_2}=\widehat{A_2}\)

Vì \(\widehat{A_2}=\widehat{A_3}\) nên Az là tia phân giác \(\widehat{CAx}\)

Bài làm:

Ta có: \(\hept{\begin{cases}\left(x-3,5\right)^2\ge0\\\left(y-\frac{1}{10}\right)^2\ge0\end{cases}\left(\forall x,y\right)}\)

=> \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^2\ge0\left(\forall x,y\right)\) , mà theo đề bài:

\(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^2\le0\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x-3,5\right)^2=0\\\left(y-\frac{1}{10}\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{7}{2}\\y=\frac{1}{10}\end{cases}}\)

Ta có : 

\(\left(x-3,5\right)^2\ge0\forall x\)     

\(\left(y-\frac{1}{10}\right)^4\ge0\forall y\)

Bài làm:

Ta có: \(A=\frac{1}{8.14}+\frac{1}{14.20}+...+\frac{1}{50.56}\)

\(A=\frac{1}{6}\left(\frac{6}{8.14}+\frac{6}{14.20}+...+\frac{6}{50.56}\right)\)

\(A=\frac{1}{6}\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+...+\frac{1}{50}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}\left(\frac{1}{8}-\frac{1}{50}\right)\)

\(A=\frac{1}{6}\cdot\frac{21}{200}=\frac{21}{1200}\)

:)))))))

Bg

Ta có: S = \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

=> S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

=> S = \(\frac{1}{5}-\frac{1}{95}\)

=> S = \(\frac{19}{95}-\frac{1}{95}\)

=> S = \(\frac{18}{95}\)

Vậy S = \(\frac{18}{95}\)

\(S=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{93\cdot95}\)

\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(S=\left(\frac{1}{5}-\frac{1}{95}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{93}+\frac{1}{93}\right)\)

\(S=\left(\frac{1}{5}-\frac{1}{95}\right)\)

\(S=\frac{19}{95}-\frac{1}{95}\)

\(S=\frac{18}{95}\)

\(\frac{...}{21}-\frac{2}{3}=\frac{5}{21}=>\frac{...}{21}=\frac{5}{21}+\frac{2}{3}=>\frac{...}{21}=\frac{19}{21}\)

học tốt

\(=\frac{19}{21}\)

\(2^3+3.\left(\frac{1}{9}\right)^0-2^{-2}.4+\left[\left(-2\right)^2\div\frac{1}{2}.8\right]\)

\(=8+3.1-\frac{1}{2^2}.4+\left[4\div\frac{1}{2}.8\right]\)

\(=8+3-1+64\)

\(=74\)

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