Số số hạng là: [(2n-1) - 1] : 2 +1 = n (số hạng)

M = n(2n-1+1) : 2 = n(2n):2 = n²

=> M là số chính phương

\(\left(2\cdot x-5\right)^5-64=960\\ \left(2x-5\right)^5=1024\\ \left(2x-5\right)^5=4^5\\ 2x-5=4\\ 2x=9\\ x=\dfrac{9}{2}\)

a.x+(x+10)+(x+14)+....+(x+162)=3434

x+(x+x+.....+x)+(10+14+18+.....+162)=3434

x+x.39+3354=3434

x⋅40=80

x=2

\(a.x+\left(x+10\right)+\left(x+14\right)+....+\left(x+162\right)=3434\\ x+\left(x+x+.....+x\right)+\left(10+14+18+.....+162\right)=3434\\ x+x+39+3354=3434\\ x\cdot40=80\\ x=2\)

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)

=> 2x=98

=> x=49

\(2^{x+1}-1=63\\ 2^{x+1}=64\\ 2^{x+1}=2^6\\ =>X+1=6\\ =>x=5\)

\(\left(X-3\right)\cdot4^5=4^8\\ X-3=64\\ =>X=67\)

\(\left(X-3\right)\cdot4^5=4^8\\ X-3=4^8:4^5\\ X-3=64\\ X=67\)

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

\(2\cdot3^x-1=53\\ 2\cdot3^x=54\\ 3^x=27\\ 3^x=3^3=>x=3\)

\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)

\(\dfrac{1}{6}+x=\dfrac{14}{5}\)

\(=>x=\dfrac{79}{30}\)

\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)

\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)

\(\dfrac{32}{35}:x=\dfrac{11}{6}\)

\(x=\dfrac{192}{385}\)

\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)

\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)

\(x:\dfrac{3}{4}=\dfrac{38}{30}\)

\(=>x=\dfrac{19}{20}\)