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\(\frac{3}{3.6.9}+\frac{3}{6.9.12}+\frac{3}{9.12.15}+\frac{3}{12.15.18}=\frac{3}{6}\left(\frac{6}{3.6.9}+\frac{6}{6.9.12}+\frac{6}{9.12.15}+\frac{6}{12.15.18}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{6.9}+\frac{1}{6.9}-\frac{1}{9.12}+\frac{1}{9.12}-\frac{1}{12.15}+\frac{1}{12.15}-\frac{1}{15.18}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{15.18}\right)=\frac{1}{2}.\frac{14}{270}=\frac{7}{270}\)
\(a)\)\(6-\left(15+15\right)=x-\left(15-6\right)\)
\(\Leftrightarrow6-30=x-9\)
\(\Leftrightarrow-24=x-9\)
\(\Leftrightarrow x=-24+9\)
\(\Leftrightarrow x=-15\)
Vậy\(x=-15\)
\(b)\)\(x+\frac{4}{12}=\frac{3}{-9}\)
\(\Leftrightarrow x+\frac{1}{3}=-\frac{1}{3}\)
\(\Leftrightarrow x=-\frac{1}{3}-\frac{1}{3}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy\(x=\frac{-2}{3}\)
Linz
a, \(6-\left(15+15\right)=x-\left(15-6\right)\)
\(\Leftrightarrow6-30=x-9\Leftrightarrow15=x+30\Leftrightarrow x=-15\)
b, \(\frac{x+4}{12}=\frac{3}{-9}\)
\(\Leftrightarrow-9x-36=36\Leftrightarrow-9x=72\Leftrightarrow x=-8\)
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
\(=\dfrac{-15}{4}.\dfrac{9}{11}+\dfrac{15}{4}.\dfrac{15}{11}-\dfrac{15}{4}.\dfrac{6}{11}\)
= \(\dfrac{15}{4}.\dfrac{-9}{11}+\dfrac{15}{4}.\dfrac{15}{11}-\dfrac{15}{4}.\dfrac{6}{11}\)
= \(\dfrac{15}{4}.\left(\dfrac{-9}{11}+\dfrac{15}{11}-\dfrac{6}{11}\right)\)
= \(\dfrac{15}{4}.\left(\dfrac{6}{11}-\dfrac{6}{11}\right)\)
= \(\dfrac{15}{4}.\left(\dfrac{0}{11}\right)\)
= \(\dfrac{15}{4}.0\)
= 0
\(\dfrac{1}{15}-\dfrac{2}{15}+\dfrac{3}{15}-\dfrac{4}{15}+\dfrac{5}{15}-\dfrac{6}{15}+\dfrac{7}{15}-\dfrac{8}{15}+\dfrac{9}{15}\)
\(=\dfrac{1-2+3-4+5-6+7-8+9}{15}\)
\(=\dfrac{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+9}{15}\) => có 4 cặp
\(=\dfrac{\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+9}{15}\)
\(=\dfrac{\left(-1\right).4+9}{15}\)
\(=\dfrac{\left(-4\right)+9}{15}\)
\(=\dfrac{5}{15}=\dfrac{1}{3}\)
= -6 - 15 - 9
= -21 - 9
= -30