Question

\(\frac{1}{3.7}+\frac{1}{7.11}+.....+\frac{1}{\left(4x+3\right)\left(4x+7\right)}=\frac{5}{12}\)

Answer 1

\(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{\left(4x+3\right)\left(4x+7\right)}=\frac{5}{12}\)(x phải khác \(-\frac{3}{4};-\frac{7}{4}\)nhé)

\(\Leftrightarrow\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4x+3\right)\left(4x+7\right)}=4.\frac{5}{12}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4x+3}-\frac{1}{4x+7}=\frac{5}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{4x+7}=\frac{5}{3}\)

\(\Leftrightarrow\frac{4x+7-3}{3\left(4x+7\right)}=\frac{5\left(4x+7\right)}{3\left(4x+7\right)}\)

\(\Rightarrow4x+7-3=20x+35\)(chỗ này dùng dấu suy ra nhé)

\(\Leftrightarrow4x-20x=35-7+3\)

\(\Leftrightarrow-16x=31\)

\(\Leftrightarrow x=-\frac{31}{16}\)

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