Question

Tìm x \(\in\) Z :

a \(\left(\dfrac{3}{4}\right)^{2x}=\dfrac{12.5^9.\left(-7\right)^{10}+3.7^9.\left(-5\right)^{11}}{35^9.2^4}\)

b, \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

NHANH NHÁ GIÚP MINK VỚI GIẢI RÕ CHI TIẾT MAI MIK NỘP RỒI ! THANKS TRC^^

Answer 1

b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

Answer 2

mik lỡ bấm nhầm rồi, phần sau bn tự nghĩ nhé, sorry

Answer 3

b)\(\dfrac{2}{6}\)+\(\dfrac{2}{12}\)+.......+\(\dfrac{2}{x.\left(x+1\right)}\)=\(\dfrac{499}{1000}\)

\(\dfrac{2}{2.3}\)+\(\dfrac{2}{3.4}\)+.........+\(\dfrac{2}{x.\left(x+1\right)}\)=\(\dfrac{499}{1000}\)

2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.........+\(\dfrac{1}{x.\left(x+1\right)}\))=\(\dfrac{499}{1000}\)

2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-..........+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))=\(\dfrac{499}{1000}\)

2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))=\(\dfrac{499}{1000}\)

\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{499}{1000}\):2

\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{499}{2000}\)

\(\dfrac{1}{x+1}\)=\(\dfrac{1}{2}\)-\(\dfrac{499}{2000}\)

\(\dfrac{1}{x+1}\)=\(\dfrac{501}{2000}\)

Đề bài sai phải

Answer 4

Quang tèo

Answer 5

Đề bài câu b sai phải bạn ạ vô lí