giải chi tiết giúp mình nhé

Tính chất nếu: 

\(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) 

Ta có:

\(A=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}\)

\(A>\dfrac{10^{99}+10}{10^{89}+10}\)

\(A>\dfrac{10\cdot\left(10^{98}+1\right)}{10\cdot\left(10^{88}+1\right)}\)

\(A>\dfrac{10^{98}+1}{10^{88}+1}\)

\(A>B\)

\(A=\dfrac{10^{99}+1}{10^{89}+1}< \dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{89}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}\)

Vậy \(A< B\)

a) 

11*3¹¹ = 1948617

3¹³ = 1594323

vì 1948617 > 1594323 nên 11*3¹¹ > 3¹³

b) 89² = 7921

88*90 = 7920

vì 7921 > 7920 nên 89² > 88*90

mình cảm ơn :3

a) Ta có :

\(\frac{7}{9}< 1\); \(\frac{19}{17}>1\)

Vì \(\frac{7}{9}< 1< \frac{19}{17}\)nên \(\frac{7}{9}< \frac{19}{17}\)

b) Xét phân số trung gian là \(\frac{n}{n+2}\)

Vì \(\frac{n}{n+3}< \frac{n}{n+2}\)và \(\frac{n}{n+2}< \frac{n+1}{n+2}\)

\(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)

c) Ta có :

\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)

Vậy \(A< B\)

a)7/9<1,19/17 => 7/9<19/17.

Anh cũng nằm trong đội tuyển nàk em tham khảo nhé 

Ta có : 

\(A=\frac{10^{11}-1}{10^{12}-1}\)

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

Lại có : 

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

\(\Leftrightarrow\)\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)\(\left(2\right)\)

Từ (1) và (2) suy ra \(10A< 1< 10B\) hay \(A< B\)

Vậy \(A< B\)

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

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

Sao sánh 10A với 10B 

Vì 1=1 nên so sánh \(-\frac{9}{10^{12}-1}\)với \(\frac{9}{10^{11}+1}\)

=> \(-\frac{9}{10^{12}-1}< \frac{9}{10^{11}+1}\)

=> 10A < 10B

=> A < B

ĐỀ : \(A=\frac{10^{11}-1}{10^{12}-1}\)và \(B=\frac{10^{10}+1}{10^{11}+1}\).

Ta có : \(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}\)\(=1-\frac{9}{10^{12}-1}\).

            \(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}\)\(=1+\frac{9}{10^{11}+1}\).

Ta so sánh : \(1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\)hay \(10A< 10B.\)

\(\Rightarrow A< B\left(đpcm\right)\)