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\(\frac{13}{7}-\frac{8}{6}+\frac{8}{7}-\frac{4}{6}-\frac{1}{3}\)
\(=\left(\frac{13}{7}+\frac{8}{7}\right)-\left(\frac{8}{6}+\frac{4}{6}\right)-\frac{1}{3}\)
\(=3-2-\frac{1}{3}\)
\(=1-\frac{1}{3}\)
\(=\frac{2}{3}\)
Ta có \(\frac{13}{7}-\frac{8}{6}+\frac{8}{7}-\frac{4}{6}-\frac{1}{3}\)
\(=\frac{13}{7}-\frac{8}{6}+\frac{8}{7}-\frac{4}{6}-\frac{2}{6}\)
\(=\left(\frac{13}{7}+\frac{8}{7}\right)-\left(\frac{8}{6}+\frac{4}{6}+\frac{2}{6}\right)\)
\(=3-2\)
\(=1\)
\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)
\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)
\(\rightarrow10x+80+15x+105=-6x\)
\(\Leftrightarrow31x+185=0\)
\(\Leftrightarrow x=-\frac{185}{31}\)
b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)
\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)
\(\rightarrow20x-160+15x-105=240+12-12x\)
\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)
\(\frac{5}{2}+\frac{1}{4}-\frac{1}{2}+\frac{3}{4}-\frac{7}{2}\)
\(=\left(\frac{5}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)-\frac{7}{2}\)
\(=2+1-\frac{7}{2}\)
\(=3-\frac{7}{2}\)
\(=\frac{6}{2}-\frac{7}{2}\)
\(=\frac{-1}{2}\)
Ta có: \(\left(\dfrac{4}{7}-\dfrac{1}{3}\right)^x=8\)
\(\Leftrightarrow\left(\dfrac{5}{21}\right)^x=8\)