Thu gọn phương trình sau a(3x-1)/ 5 -6x-7/4 + 3x+2/ 10= 0 Mấy bạn giúp mình làm bài này vs
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Bài 1:
\(\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+...+\frac{4}{306}\)
\(=4\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{306}\right)\)
\(=4\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{17\cdot18}\right)\)
\(=4\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}\right)\)
\(=4\cdot\left(\frac{1}{3}-\frac{1}{18}\right)\)
\(=4\cdot\left(\frac{6}{18}-\frac{1}{18}\right)\)
\(=4\cdot\frac{5}{18}\)
\(=\frac{10}{9}\)
Bài 2 :
\(\left(3x-4\right)-\left(6x+7\right)=8\)
\(3x-4-6x-7=8\)
\(\left(3x-6x\right)-\left(4+7\right)=8\)
\(-3x-11=8\)
\(-3x=8+11\)
\(-3x=19\)
\(x=19:\left(-3\right)\)
\(x=\frac{-19}{3}\)
Vậy \(x=\frac{-19}{3}\)
b ) \(\left(\frac{4}{5}x+3\right):\left(-4\right)=\frac{1}{2}\)
\(\frac{4}{5}x+3=\frac{1}{2}\cdot\left(-4\right)\)
\(\frac{4}{5}x+3=-2\)
\(\frac{4}{5}x=\left(-2\right)-3\)
\(\frac{4}{5}x=-5\)
\(x=\left(-5\right):\frac{4}{5}\)
\(x=\left(-5\right)\cdot\frac{4}{5}\)
\(x=-4\)
Vậy \(x=-4\)
k nha !
\(\frac{4}{12}\)+\(\frac{4}{20}\)+...+\(\frac{4}{306}\)=\(\frac{4}{3.4}\)+\(\frac{4}{4.5}\)+...+\(\frac{4}{17.18}\)=4(\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{17}\)-\(\frac{1}{18}\))
=4(\(\frac{1}{3}\)-\(\frac{1}{8}\))=4.\(\frac{5}{24}\)=\(\frac{5}{6}\)
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