A là trực tâm của tam giác ABC vì A là giao của 2 đường cao AB và AC

Chọn C

Xét ΔABC có

AC là đường cao ứng với cạnh AB

AB là đường cao ứng với cạnh AC

AC cắt AB tại A

Do đó: A là trực tâm của ΔABC

a: Gọi tử số của phân số cần tìm là a

Theo đề, ta có: \(\dfrac{2}{3}< \dfrac{a}{30}< \dfrac{5}{6}\)

=>\(\dfrac{20}{30}< \dfrac{a}{30}< \dfrac{25}{30}\)

=>20<a<25

Vậy: Các phân số cần tìm có dạng là \(\dfrac{a}{30};20< a< 25\)

b: Gọi mẫu số của phân số cần tìm là a

Theo đề, ta có: \(\dfrac{-5}{6}< \dfrac{-15}{a}< \dfrac{-3}{4}\)

=>\(\dfrac{5}{6}>\dfrac{15}{a}>\dfrac{3}{4}\)

=>\(\dfrac{15}{18}>\dfrac{15}{a}>\dfrac{15}{20}\)

=>18<a<20

Vậy: Các phân số cần tìm có dạng là \(-\dfrac{15}{a};18< a< 20\)

a: Xét ΔBMD và ΔCMA có

MB=MC

\(\widehat{BMD}=\widehat{CMA}\)(hai góc đối đỉnh)

MD=MA

Do đó: ΔBMD=ΔCMA
=>\(\widehat{MBD}=\widehat{MCA}\)

=>BD//AC

=>\(\widehat{ABD}+\widehat{BAC}=180^0\)

mà \(\widehat{BAC}+\widehat{B'A'C'}=180^0\)

nên \(\widehat{ABD}=\widehat{B'A'C'}\)

b: ΔBMD=ΔCMA

=>BD=CA

mà CA=A'C'

nên BD=A'C'

Xét ΔABD và ΔB'A'C' có

AB=B'A'

\(\widehat{ABD}=\widehat{B'A'C'}\)

BD=A'C'

Do đó: ΔABD=ΔB'A'C'

=>AD=B'C'

mà \(AM=\dfrac{1}{2}AD\)

nên \(AM=\dfrac{1}{2}B'C'\)

a: \(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-2\dfrac{2}{5}\)

=>\(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-\dfrac{12}{5}\)

=>\(\dfrac{1}{3}-5x=\dfrac{32}{15}:\dfrac{-12}{5}=\dfrac{32}{15}\cdot\dfrac{-5}{12}=\dfrac{-160}{180}=\dfrac{-8}{9}\)

=>\(5x=\dfrac{1}{3}+\dfrac{8}{9}=\dfrac{3}{9}+\dfrac{8}{9}=\dfrac{11}{9}\)

=>\(x=\dfrac{11}{9}:5=\dfrac{11}{45}\)

b: \(\left(2x-\dfrac{1}{3}\right)^5=\dfrac{32}{243}\)

=>\(\left(2x-\dfrac{1}{3}\right)^5=\left(\dfrac{2}{3}\right)^5\)

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

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

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

c: \(\left(\dfrac{5}{6}x+3\right)^2=\dfrac{4}{9}\)

=>\(\left[{}\begin{matrix}\dfrac{5}{6}x+3=\dfrac{2}{3}\\\dfrac{5}{6}x+3=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}x=\dfrac{2}{3}-3=-\dfrac{7}{3}\\\dfrac{5}{6}x=-\dfrac{2}{3}-3=-\dfrac{11}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{7}{3}:\dfrac{5}{6}=-\dfrac{7}{3}\cdot\dfrac{6}{5}=\dfrac{-14}{5}\\x=-\dfrac{11}{3}:\dfrac{5}{6}=-\dfrac{11}{3}\cdot\dfrac{6}{5}=\dfrac{-22}{5}\end{matrix}\right.\)

a) 

\(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-2\dfrac{2}{5}\\ \dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-\dfrac{8}{5}\\ \dfrac{1}{3}-5x=\dfrac{32}{15}:-\dfrac{8}{5}\\ \dfrac{1}{3}-5x=\dfrac{-4}{3}\\ 5x=\dfrac{1}{3}+\dfrac{4}{3}\\ 5x=\dfrac{5}{3}\\ x=\dfrac{5}{3}:5\\ x=\dfrac{1}{3}\) 

b) \(\left(2x-\dfrac{1}{3}\right)^5=\dfrac{32}{243}\) 

\(\left(2x-\dfrac{1}{3}\right)^5=\left(\dfrac{2}{3}\right)^5\)

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

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

\(2x=1\\ x=\dfrac{1}{2}\) 

c) \(\left(\dfrac{5}{6}x+3\right)^2=\dfrac{4}{9}\)

\(\left(\dfrac{5}{6}x+3\right)^2=\left(\dfrac{2}{3}\right)^2\)

TH1: 

\(\dfrac{5}{6}x+3=\dfrac{2}{3}\\ \dfrac{5}{6}x=\dfrac{2}{3}-3\\ \dfrac{5}{6}x=-\dfrac{7}{3}\\ x=\dfrac{-7}{3}:\dfrac{5}{6}=-\dfrac{14}{5}\)

TH2: 

\(\dfrac{5}{6}x+3=-\dfrac{2}{3}\\ \dfrac{5}{6}x=-\dfrac{2}{3}-3\\ \dfrac{5}{6}x=-\dfrac{11}{3}\\ x=-\dfrac{11}{3}:\dfrac{5}{6}\\ x=-\dfrac{22}{5}\)

\(2\left(x+1\dfrac{1}{3}\right)=\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{6}\\ x+\dfrac{4}{3}=\dfrac{1}{6}:2\\ x+\dfrac{4}{3}=\dfrac{1}{12}\\ x=\dfrac{1}{12}-\dfrac{4}{3}\\ x=-\dfrac{15}{12}=\dfrac{-5}{4}\)

\(2\left(x+1\dfrac{1}{3}\right)=\left(-\dfrac{1}{2}\right)^2\cdot\dfrac{2}{3}\)

=>\(2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}=\dfrac{1}{6}\)

=>\(x+\dfrac{4}{3}=\dfrac{1}{12}\)

=>\(x=\dfrac{1}{12}-\dfrac{4}{3}=\dfrac{1}{12}-\dfrac{16}{12}=-\dfrac{15}{12}=-\dfrac{5}{4}\)

\(\left(-\dfrac{3}{5}\right)^2-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\)

=>\(\dfrac{9}{25}-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\)

=>\(x-\dfrac{1}{3}=\dfrac{9}{25}-\dfrac{4}{25}=\dfrac{5}{25}=\dfrac{1}{5}\)

=>\(x=\dfrac{1}{5}+\dfrac{1}{3}=\dfrac{8}{15}\)

`x- \left(\frac54-\frac75 \right)=\frac{9}{20}`

`\Rightarrow x-\frac{-3}{20}=\frac{9}{20}`

`\Rightarrow x=\frac{9}{20}+\frac{-3}{20}`

`\Rightarrow x=\frac{3}{10}`

\(x-\left(\dfrac{5}{4}-\dfrac{7}{5}\right)=\dfrac{9}{20}\)

=>\(x-\dfrac{25-28}{20}=\dfrac{9}{20}\)

=>\(x+\dfrac{3}{20}=\dfrac{9}{20}\)

=>\(x=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)

a: \(-0,7< \dfrac{-13}{19}< -0,6\)

\(\dfrac{19}{-23}< -0,8\)

mà -0,8<-0,7

nên \(\dfrac{19}{-23}< -\dfrac{13}{19}\)

b: \(\dfrac{1}{83}:\dfrac{6}{331}=\dfrac{1}{83}\cdot\dfrac{331}{6}=\dfrac{331}{498}< 1\)

=>\(\dfrac{1}{83}< \dfrac{6}{331}\)

=>\(\dfrac{1}{83}+1< \dfrac{6}{331}+1\)

=>\(\dfrac{84}{83}< \dfrac{337}{331}\)

=>\(\dfrac{84}{-83}>\dfrac{-337}{331}\)