1/

\(10A=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 1< 10B$

$\Rightarrow A< B$

2/

\(C=\frac{10^{99}+5}{10^{99}-8}=1+\frac{13}{10^{99}-8}\)

\(D=\frac{10^{100}+6}{10^{100}-4}=1+\frac{10}{10^{100}-4}\)

So sánh \(\frac{13}{10^{99}-8}=\frac{130}{10^{100}-80}> \frac{130}{10^{100}-4}> \frac{10}{100^{100}-4}\)

$\Rightarrow 1+\frac{13}{10^{99}-8}> 1+\frac{10}{100^{10}-4}$

$\Rightarrow C> D$

ko bíb

So sánh: mk làm luôn nè:

Ta có: \(\frac{10}{11}>\frac{10}{11+12};\frac{11}{12}>\frac{11}{11+12}\)

\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10}{11+12}+\frac{11}{11+12}\)

\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10+11}{11+12}\)

MK KO BIẾT ĐÚNG KO NỮA NÊN BN CÓ THỂ THAM KHẢO CỦA CÁC BẠN KHÁC NHÉ.!!

CHÚC BẠN HỌC TỐT. ^_^

 1/2+1/2^2+1/3^2+1/4^2+.......+1/100^2<1

= 1/2 + 1/4 + 1/9 + ... + 1/10000

có : 100 - 1 + 1 = 100 số hạng 

1 = 1/100 + 1/100 + ... + 1/100

suy ra  1/2+1/2^2+1/3^2+1/4^2+.......+1/100^2<1

\(\dfrac{11}{10}< \dfrac{9}{30}\)

\(\dfrac{6}{7}>\dfrac{3}{5}\)

\(\dfrac{25}{100}< \dfrac{3}{4}\)

a \(\dfrac{11}{10}>\dfrac{3}{10}\)

b \(\dfrac{30}{35}>\dfrac{21}{35}\)

c \(\dfrac{1}{4}< \dfrac{3}{4}\)

a, \(\dfrac{11}{10}< \dfrac{9}{30}\)

b, \(\dfrac{6}{7}>\dfrac{3}{5}\)

c, \(\dfrac{25}{100}< \dfrac{3}{4}\)

s2 Lắc Lư s2 cko hỏi ôg lp mấy z? 

2/

a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)

\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)

-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)

b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)

-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)

c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)

-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)

a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)

c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)

a) 

b) 

c) \(\dfrac{25}{100}=\dfrac{10}{40}\)