C1:A = \(\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}\)
= \(1+\frac{3}{10^{50}-1}\)
B = \(\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}\)
= \(1+\frac{3}{10^{50}-3}\)
Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)=) \(1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\)=) \(A< B\)
C2: Áp dụng tính chất : Nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Vì B > 1 =) B > \(\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
(=) B > A

Ta có:

\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}\)

Vì \(10^{50}-1>10^{50}-3\Rightarrow\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)(2 phân số có cùng tử số, mẫu số của phân số nào lớn hơn thì phân  

                                                                                             số đó nhỏ hơn)

\(\Rightarrow1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\Rightarrow A< B\)     

\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}.\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}.\)

Do 10⁵⁰-1 > 10⁵⁰-3 ; => \(1+\frac{3}{10^{50}-3}>1+\frac{3}{10^{50}-1}\)

=> B > A

$3^{20}$ và ${27}^4$

Ta có: ${27}^4=(3^3{)}^4=3^{12}<3^{20}$

$\Rightarrow 3^{20}>{27}^4$

$2^{25}$ và ${16}^6$

Ta có:

${16}^6=(2^4{)}^6=2^{24}<2^{25}$

$\Rightarrow 2^{25}>2^{24}$

lưu ý 27 mà số 4 nhỏ ấy đấy là mũ nhé nhầm

a, Ta có: 5^30 = (5^3)^10= 125^ 10    >   (-10^2)^10= 100^10

b, ta có: 21^12= ( 21^3)^4 > 54^4

 c, Ta có: (1/16)^10 = 1/16^10

                 (1/2)^50= 1/2^50

   Lại có: 16^10=(2^4)^10= 2^40 < 2^50 nên (1/6)^10> (1/2)^50

Đặt \(A=1+6+6^2+\cdots+6^9\)

=>\(6A=6+6^2+6^3+\cdots+6^{10}\)

=>6A-A=\(6+6^2+6^3+\cdots+6^{10}-1-6-6^2-\cdots-6^9\)

=>5A=\(6^{10}-1\)

=>\(A=\frac{6^{10}-1}{5}\)

\(P=\frac{6^{10}}{1+6+6^2+\cdots+6^9}\)

\(=\frac{6^{10}}{\frac{6^{10}-1}{5}}=\frac{5\cdot6^{10}}{6^{10}-1}=\frac{5\cdot6^{10}-5+5}{6^{10}-1}=5+\frac{5}{6^{10}-1}\)

=>P<6

\(B=1+8+8^2+\cdots+8^9\)

=>\(8B=8+8^2+\cdots+8^{10}\)

=>8B-B=\(8+8^2+\cdots+8^{10}-1-8-\cdots-8^9\)

=>7B=\(8^{10}-1\)

=>\(B=\frac{8^{10}-1}{7}\)

\(Q=\frac{8^{10}}{1+8+8^2+\cdots+8^9}\)

\(=8^{10}:\frac{8^{10}-1}{7}=\frac{8^{10}\cdot7}{8^{10}-1}=\frac{7\cdot8^{10}-7+7}{8^{10}-1}=7+\frac{7}{8^{10}-1}\)

=>Q>7>6

=>Q>P

\(A=\frac{10^{50}+2}{10^{50}+1}=\frac{2}{1}=2\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{-1}{3}\)

\(\Rightarrow A>B\)

Ta có: \(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}=1+\frac{3}{10^{50}-1}\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}=1+\frac{3}{10^{50}-3}\)

Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\Rightarrow1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\Rightarrow A< B\)