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16:Sửa đề: \(\frac12x+\frac16\left(x-2\right)=\frac34-2x\)
=>\(\frac12x+\frac16x-\frac26=\frac34-2x\)
=>\(\frac46x-\frac13=\frac34-2x\)
=>\(\frac23x+2x=\frac34+\frac13\)
=>\(\frac83x=\frac{9}{12}+\frac{4}{12}=\frac{13}{12}\)
=>\(x=\frac{13}{12}:\frac83=\frac{13}{12}\times\frac38=\frac{13}{4\times8}=\frac{13}{32}\)
19: \(\frac{5}{12}x+3=\frac13-\frac{7}{12}x\)
=>\(\frac{5}{12}x+\frac{7}{12}x=\frac13-3\)
=>\(\frac{12}{12}x=\frac13-\frac93=-\frac83\)
=>\(x=-\frac83\)
20: \(\frac12x+\frac52=\frac72x-\frac34\)
=>\(\frac12x-\frac72x=-\frac34-\frac52\)
=>\(-3x=-\frac34-\frac{10}{4}=-\frac{13}{4}\)
=>\(3x=\frac{13}{4}\)
=>\(x=\frac{13}{4}:3=\frac{13}{12}\)
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{x+1}{2}\)
\(=\frac{1}{2}\left(2+3+4+...+x+1\right)\)
\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)
\(=\frac{1}{2}\cdot\frac{x\left(x+3\right)}{2}=\frac{x\left(x+3\right)}{4}\).