\(AM=\frac{BC}{2}\Rightarrow AM=BM=CM\)

=> tg ABM cân tại M \(\Rightarrow\widehat{ABC}=\widehat{BAM}\)

Và tg ACM cân tại M \(\Rightarrow\widehat{ACB}=\widehat{CAM}\)

\(\Rightarrow\widehat{ABC}+\widehat{ACB}=\widehat{BAM}+\widehat{CAM}=\widehat{BAC}\)

Mà \(\widehat{ABC}+\widehat{ACB}=180^o-\widehat{BAC}=\widehat{BAC}\Rightarrow\widehat{BAC}=90^o\)

=> tg ABC vuông tại A

a)ta có \(\frac{B}{2^3}=\frac{2^3+4^3+6^3+...+20^3}{2^3}\)

            \(\frac{B}{2^3}=1^3+2^3+3^3+...+10^3\) mà 1³+2³+3³+...+10³=3025

      nên\(\frac{B}{2^3}=3025\)

             \(\Rightarrow B=378,125\)

a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)

Tự làm nốt và kết luận 

b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)

Vậy ....

a/ \(\left|\frac{5}{8}\right|< \left|\frac{5}{12}\right|\Rightarrow\frac{-5}{8}>\frac{-5}{12}\)

\(\left|\frac{5}{12}\right|< \left|\frac{7}{12}\right|\Rightarrow\frac{-5}{12}>\frac{7}{-12}\)

\(\Rightarrow\frac{-5}{8}>\frac{7}{-12}\)

b/

\(\frac{300}{-4141}=\frac{300}{-41.101}\)

Ta có \(\left|\frac{300}{41.101}\right|< \left|\frac{300}{41.100}\right|=\left|\frac{3}{41}\right|\)

\(\Rightarrow\frac{300}{-4141}=\frac{300}{-41.101}>\frac{-3}{41}\)

x là chữ số chưa biết hay bạn viết nhầm