hơi dài á

a, ta có:

kẻ đường cao CH

=> S_ABC \(=\frac{CHxAB}{2}\) ; S_CDB \(=\frac{BDxCH}{2}\)

mà BD= \(\frac{1}{4}\)AB nên S_CDB=\(\frac{1}{4}\)xS_ABC

CMTT với S_BEC = \(\frac{1}{4}\)xS_ABC

_{b, ta có:} S_CDB = S_BEC =  \(\frac{1}{4}\)xS_ABC

_=> S_DBG+S_BGC = S_BGC+S_EGC

_=> S_DBG=S_EGC

ccccgggggggggg

\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+49\)

\(=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+...+\left(\frac{48}{2}+1\right)+1\)

\(=\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+1\)

\(=\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...\frac{50}{2}\)

\(=50.\left(\frac{1}{50}+\frac{1}{49}+...\frac{1}{2}\right)\)

\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)}\)

\(\Rightarrow\frac{S}{P}=\frac{1}{50}\)

<br class="Apple-interchange-newline"><div></div>P=149 +248 +347 +...+482 +49

=(149 +1)+(248 +1)+...+(482 +1)+1

=(5049 +5048 +...+502 )+1

=5050 +5049 +5048 +...502 

=50.(150 +149 +...12 )

\(\frac{2}{5}-\frac{x}{7}=\frac{1}{4}+\frac{2}{-9}=\frac{-1}{36}\)

\(\frac{x}{7}=\frac{2}{5}-\frac{-1}{36}=\frac{2}{5}+\frac{1}{36}=\frac{67}{180}\)

#Monster

bạn tự làm tiếp nha

\(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2\left(\frac{2019}{4020}\right)=\frac{673}{670}\)

Ta có \(\frac{1}{10.11}=\frac{1}{10}-\frac{1}{11}\)

... \(\frac{1}{69.70}=\frac{1}{69}-\frac{1}{70}\)

Vì vậy

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+..+\frac{1}{69}-\frac{1}{70}\right)=7\left(\frac{1}{10}-\frac{1}{70}\right)=\frac{3}{5}\)

a)$\widehat{MOy}={180}^{\circ }-\widehat{xOM}=180-60={120}^{\circ }$

⇒$\widehat{MON}=\widehat{yON}-\widehat{MOy}=150-120={30}^{\circ }$

b)Ta có: $\widehat{xON}=\widehat{xOM}-\widehat{MON}=60-30={30}^{\circ }$

⇒$\widehat{xON}=\widehat{MON}$

⇒ON là tia phân giác của $\widehat{xOM}$

Tham khảo : Câu hỏi của Hương Chi - Toán lớp 6 - Học trực tuyến OLM

\(\frac{7}{-25}+\frac{-18}{25}+\frac{4}{23}+\frac{5}{7}+\frac{19}{23}\)   

\(=\left(-\frac{7}{25}-\frac{18}{25}\right)+\left(\frac{4}{23}+\frac{19}{23}\right)+\frac{5}{7}\)   

\(=-1+1+\frac{5}{7}\)   

\(=0+\frac{5}{7}\)   

\(=\frac{5}{7}\)

thank