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

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

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

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

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

\(=74\)

bạn ơi thiếu đề

 \(\frac{-11}{14}-\frac{-4}{...}=\frac{-3}{14}\)

\(\frac{-11}{14}+\frac{4}{...}=\frac{-3}{14}\)

\(\frac{4}{...}=\frac{-3}{14}-\frac{-11}{14}\)

\(\frac{4}{...}=\frac{-3}{14}+\frac{11}{14}\)

\(\frac{4}{...}=\frac{8}{14}\)

\(\frac{4}{...}=\frac{4}{7}\)

\(\frac{1}{...}-\frac{-2}{15}=\frac{7}{15}\)

\(\frac{1}{...}+\frac{2}{15}=\frac{7}{15}\)

\(\frac{1}{...}=\frac{7}{15}-\frac{2}{15}\)

\(\frac{1}{...}=\frac{5}{15}\)

\(\frac{1}{...}=\frac{1}{3}\)

Hok tốt !!!!!!!!!

Bài làm:

Ta có: \(\frac{a}{b}=\frac{3}{4}\Leftrightarrow\frac{a}{3}=\frac{b}{4}\) 

Áp dụng t/c dãy tỉ số bằng nhau: 

\(\frac{a}{3}=\frac{b}{4}=\frac{-3a+5b}{-9+20}=\frac{33}{11}=3\)

=> \(\hept{\begin{cases}a=33\\b=44\end{cases}}\)

-3a + 5b = 33 

=> -( 3a - 5b ) = 33

=> 3a - 5b = -33

\(\hept{\begin{cases}\frac{a}{b}=\frac{3}{4}\\3a-5b=-33\end{cases}}\Rightarrow\hept{\begin{cases}\frac{a}{3}=\frac{b}{4}\\3a-5b=-33\end{cases}\Rightarrow}\hept{\begin{cases}\frac{3a}{9}=\frac{5b}{20}\\3a-5b=-33\end{cases}}\)

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

\(\frac{3a}{9}=\frac{5b}{20}=\frac{3a-5b}{9-20}=\frac{-33}{-11}=3\)

\(\Rightarrow\hept{\begin{cases}\frac{3a}{9}=3\Rightarrow a=9\\\frac{5b}{20}=3\Rightarrow b=12\end{cases}}\)

Az // BC => \(\widehat{xAz}=\widehat{ABC}\)(đồng vị) (1)
Az // BC => \(\widehat{CAz}=\widehat{ACB}\)(so le trong) (2)
Từ (1), (2) và \(\widehat{ABC}=\widehat{ACB}\)=> \(\widehat{xAz}=\widehat{CAz}\)
=> Az là tia phân giác của góc CAx.
Vì không nhớ cách làm chi tiết nên chị viết tắt nhé.

\(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}\)

\(=\frac{1\left(2^5+2^6+2^7+2^8\right)}{2^4\left(2^5+2^6+2^7+2^8\right)}\)

\(=\frac{1}{2^4}=\frac{1}{16}\)

Ta có \(\frac{1}{16}< \frac{1}{6}\)

=> \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}< \frac{1}{6}\)

So sánh \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}\) với \(\frac{1}{6}\) ?

Ta có: \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}=\frac{2^5.\left(1+2+2^2+2^3\right)}{2^9.\left(1+2+2^2+2^3\right)}\)

\(=\frac{1}{2^4}=\frac{1}{16}< \frac{1}{6}\)

Vậy \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}< \frac{1}{6}\)