là bài nào vậy chỉ rõ cho mik

\(\dfrac{14}{15}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{14}{15}-\dfrac{4}{15}.\dfrac{5}{2}=\dfrac{14}{15}-\dfrac{2}{3}=\dfrac{14}{15}-\dfrac{10}{15}=\dfrac{4}{15}\)

lm bài nào vậy bạn ???

Thể tích của mực nuớc biển là

\(60cm.40cm.25cm=60000cm^3=600dm^3\)

đ/s

\(a,2^2.3^2-5.2.3=4.9-5.2.3=36-30=6\)

\(b,5^2.2+20:2^2=25.2+20:4=50+5=55\)

\(c,7^2.15-5.7^2=49.15-5.49=49.\left(15-5\right)=49.10=490\)

\(d,3.5^2+15.2^2-1^2.3=3.25+15.4-3=75+60-1=134\)

\(g,425.7.4-1700.6=425.4.7-1700.6=1700.7-1700.6=1700\left(7-6\right)=1700.1=1700\)

\(h,52.39+47.39+39=\left(52+47+1\right).39=100.39=3900\)

\(i,3.51.16+4.12.87-2.24.38=3.16.51+4.12.87-2.24.38=48.51+48.87-48.38=48.\left(51+87-38\right)=48.100=4800\)

\(k,6.15.3+49.2.9+36.18=15.6.3+49.2.9+36.18=15.18+49.18+36.18=\left(15+49+36\right).18=100.18=1800\)

\(l,54.37+46.37-54.19-46.19=\left(54+46\right).37-\left(54-46\right).19=100.37-8.19=3700-152=3548\)

\(m,5.7.77-7.60+49.25-15.42=2695-420+74-630=1719\)

\(b,\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2=\left(\dfrac{-1}{12}\right)^2=\dfrac{1}{144}\)

\(c,2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^3=2:\left(\dfrac{-1}{6}\right)^3=2:\dfrac{-1}{216}=2.\left(-216\right)=-432\)

\(f,\left(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{5}{6}\right).\left(\dfrac{-3}{2}\right)^2=\left(\dfrac{3}{6}-\dfrac{4}{6}+\dfrac{5}{6}\right).\dfrac{9}{4}=\dfrac{2}{4}.\dfrac{9}{4}=\dfrac{9}{8}\)

\(e,\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right).\left(\dfrac{16}{20}-\dfrac{15}{20}\right)^2=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)

a, Xét △ABM và △ACM có

AM chung

AB = AC

BM = CM

⇒△ABM = △ACM (c.c.c)

b, THeo phần a, △ABM = △ACM

⇒\(\widehat{AMB}=\widehat{AMC}\) (2 góc tuơng ứng)

Lại có : \(\widehat{AMB}+\widehat{AMC}=180\) (vì góc BMC là góc bẹt)

\(\Rightarrow\widehat{AMB}=\widehat{AMC}=\dfrac{180}{2}=90độ\)

\(\Rightarrow AM\perp BC\)