D = \(\frac{1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-...-\frac{1}{79.81}\)

\(=\frac{1}{54}-\left(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{79.81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{79.81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{79}-\frac{1}{81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\frac{80}{81}\)

\(=\frac{1}{54}-\frac{40}{27}\)

\(=\frac{1}{54}-\frac{80}{54}\)

\(=\frac{79}{54}\)

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

Câu b sửa lại thành chứng tỏ AH  ⊥  At

a, Ta có: xOy = xAt = 50^o​ 

Mà 2 góc này nằm ở vị trí đồng vị

=> At // Oy

b, Vì AH ⊥ Oy 

         At // Oy

=> AH ⊥ At

c, Vì AH  ⊥ Oy 

=> OHA = 90^o 

Xét ΔOAH có: OAH + AOH + OHA = 180^o

=> OAH + 50^o + 90^o  = 180^o

=> OAH = 40^o 

tao đéo biết

Bạn tham khảo !

https://olm.vn/hoi-dap/detail/196616797378.html

Hoặc bạn bấm vào câu hỏi tương tự se thấy

=(1/2) mu 15*(1/2 mu 2) mu 20

=1/2 mu 15*1/2 mu 40

=1/2 mu 55

Dung thi k ung ho minh nha!Thanks!!!

\(=\frac{\left(-1\right)^7}{3^7}.3^7+\left(\frac{1}{8}\right)^3.8^3+\left(\frac{1}{5}\right)^3.\left(2.5\right)^3\)

\(=\left(-1\right)^7+\frac{1^3}{8^3}.8^3+\frac{1^3}{5^3}.2^3.5^3\)

\(=-1+1+2^3=2^3=8\)

Bài làm

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=40\)

\(x.\frac{1}{30}+1=0\)

\(x\frac{1}{3}=-1\)

\(x=-1:\frac{1}{3}\)

\(x=-1.3\)

\(x=-3\)

# Học tốt #

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.\)

\(=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1\)

\(=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1\)

\(=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1\)

\(=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1\)

\(=x\left(\frac{5+3-8}{30}\right)+1\)

\(=x\cdot0+1=1\)

\(\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1\)

trả lời:

  \frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.61​x+101​x−154​x+1+0.

=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1=(61​x+101​x−154​x)+1

=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=x(61​+101​−154​)+1

=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1=x(301⋅5​+301⋅3​−304⋅2​)+1

=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=x(305​+303​−308​)+1

=x\left(\frac{5+3-8}{30}\right)+1=x(305+3−8​)+1

=x\cdot0+1=1=x⋅0+1=1

\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1⇒61​x+101​x−154​x+1+0=1

ko chắc chúc bạn học tốt.