a: \(A=1+3+3^2+\cdots+3^{100}\)

=>\(3A=3+3^2+3^3+\cdots+3^{101}\)

=>3A-A=\(3+3^2+3^3+\cdots+3^{101}-1-3-\cdots-3^{100}\)

=>2A=\(3^{101}-1\)

=>\(A=\frac{3^{101}-1}{2}\)

b: \(B=2^{100}-2^{99}+2^{98}-2^{97}+\cdots+2^2-2+1\)

=>\(2B=2^{101}-2^{100}+2^{99}-2^{98}+\cdots+2^3-2^2+2\)

=>2B+B=\(2^{101}-2^{100}+2^{99}-2^{98}+\cdots-2^2+2+2^{100}-2^{99}+2^{98}-2^{97}+\cdots+2^2-2+1\)

=>3B=\(2^{101}+1\)

=>\(B=\frac{2^{101}+1}{3}\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)

\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)

\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)

\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)

\(A=\frac{x^2+2004x+2003}{x^2-x}\)

  100.10.2.5

= (10.10).10.(2.5)

=10.10.10.10

= 10^4 (10 mũ 4 nhé )

C1: 100.10.2.5=10².10.10=10⁴

C2: 100.10.2.5=100.10.10=100.100=100²

Chúc bạn học tốt!

-14 nha bạn

Ta co:

\(\frac{5+a}{8a}\)= \(\frac{-1}{12}\)

<=> 12(5+a)=-8a

<=>60+12a=-8a

<=>-20a=60

<=>a=-3

nhớ mik nha:)

theo bài ra ta có \(\frac{a+5}{8a}\)=\(\frac{-1}{12}\)suy ra (5+a).12=-8a suy ra 60+12a+8a=0 suy ra 20a=-60 suy ra a=-3

vậy..............

3/5;2/7;3/7;5/7

đúng thì k

cảm ơn

Phân số sau khi rút gọn là :

\(\frac{2}{7}\);\(\frac{5}{7}\);\(\frac{3}{5}\);\(\frac{3}{7}\).

Sắp xếp theo thứ tự tăng dần là :

\(\frac{2}{7}\);\(\frac{3}{7}\);\(\frac{3}{5}\);\(\frac{5}{7}\).

Kik mik nha ! Hok tốt !

\(\frac{1212}{3131}=\frac{1212:101}{3131:101}=\frac{12}{31}\)

12/31 nhé