\(\dfrac{1}{-12}+\dfrac{3}{4}+\dfrac{-1}{12}-\dfrac{1}{3}\)
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1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.4}=-6\)
\(\dfrac{-1}{7}-\dfrac{1}{8}=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\\ \dfrac{-15}{48}-\dfrac{1}{12}=\dfrac{-5}{16}-\dfrac{1}{12}=\dfrac{-15}{48}-\dfrac{4}{48}=\dfrac{-19}{48}\\ \dfrac{-3}{4}-\dfrac{4}{5}=\dfrac{-15}{20}-\dfrac{16}{20}=\dfrac{-31}{20}\\ \dfrac{3}{4}+\dfrac{-5}{6}-\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}-\dfrac{11}{12}=\dfrac{-12}{12}=-1\)
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
e: \(D=\dfrac{-10}{12}-\dfrac{7}{12}-\dfrac{4}{12}=\dfrac{-21}{12}=-\dfrac{7}{4}\)
f: \(F=\dfrac{-27}{36}+\dfrac{12}{36}+\dfrac{10}{36}=\dfrac{-5}{36}\)
g: \(G=\dfrac{209}{99}+\dfrac{36}{99}+\dfrac{66}{99}=\dfrac{311}{99}\)
h: \(H=\dfrac{10}{24}-\dfrac{42}{24}+\dfrac{3}{24}=-\dfrac{29}{24}\)
\(\dfrac{1}{-12}+\dfrac{3}{4}+\dfrac{-1}{12}-\dfrac{1}{3}=\dfrac{-1}{12}+\dfrac{3}{4}+\dfrac{-1}{12}-\dfrac{1}{3}=\dfrac{-1}{12}+\dfrac{9}{12}+\dfrac{-1}{12}-\dfrac{4}{12}=\dfrac{\text{3}}{12}=\dfrac{1}{4}\)
\(=\dfrac{1}{-12}+\dfrac{9}{12}+\dfrac{-1}{12}-\dfrac{4}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)