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\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)(đã sửa)
Đặt \(A=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{2}{3}\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\right)\)
\(\dfrac{2}{3}A=\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}:\dfrac{2}{3}=\dfrac{56}{305}\cdot\dfrac{3}{2}=\dfrac{84}{305}\)
\(B=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{20}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{19}{20}\)
\(B=\frac{1}{20}\)
\(15.\left(x+2\right)-12.\left(x-5\right)+21x=210\)
\(15x+30-12x+60+21x=210\)
\(24x+90=210\)
\(24x=120\)
\(x=5\)