A

A

\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}=1\frac{9}{11}\)

=>\(\left\{1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}\right\}\times\frac{1}{2}=1\frac{9}{11}\times\frac{1}{2}\)

=>\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)

=>\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{x}-\frac{1}{x+1}=\frac{10}{11}\)

=>\(1-\frac{1}{x+1}=\frac{10}{11}\)

=> \(\frac{1}{x+1}=1-\frac{10}{11}\)

=> \(\frac{1}{x+1}=\frac{1}{11}\)

=> x + 1 = 11

=> x = 10

Nhấn đúng cho mk nha^^

=  1 thôi 

Vì 1 mũ mấy chẳng bằng 1

i0 =6.5

ω=120π

t=o có giá trị lớn nhất↔i=i0 ↔vị trí biên dương→φ=0

CHỌN C