Giải:

a)Ta có:

C=1957/2007=1957+50-50/2007

                      =2007-50/2007

                      =2007/2007-50/2007

                      =1-50/2007

D=1935/1985=1935+50-50/1985

                      =1985-50/1985

                      =1985/1985-50/1985

                      =1-50/1985

Vì 50/2007<50/1985 nên -50/2007>-50/1985

⇒C>D

b)Ta có:

A=2016²⁰¹⁶+2/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1/2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=1+3/2016²⁰¹⁶-1

Tương tự: B=2016²⁰¹⁶/2016²⁰¹⁶-3

                 B=1+3/2016²⁰¹⁶-3

Vì 2016²⁰¹⁶-1>2016²⁰¹⁶-3 nên 3/2016²⁰¹⁶-1<3/2016²⁰¹⁶-3

⇒A<B

Chúc bạn học tốt!

Làm tiếp:

c)Ta có:

M=10²⁰¹⁸+1/10²⁰¹⁹+1

10M=10.(10²⁰¹⁸+1)/20²⁰¹⁹+1

10M=10²⁰¹⁹+10/10²⁰¹⁹+1

10M=10²⁰¹⁹+1+9/10²⁰¹⁹+1

10M=10²⁰¹⁹+1/10²⁰¹⁹+1 + 9/10²⁰¹⁹+1

10M=1+9/10²⁰¹⁹+1

Tương tự:

N=10²⁰¹⁹+1/10²⁰²⁰+1

10N=1+9/10²⁰²⁰+1

Vì 9/10²⁰¹⁹+1>9/10²⁰²⁰+1 nên 10M>10N

⇒M>N

Chúc bạn học tốt!

con cặc

Lời giải:

a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)

Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$

Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$

Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$

b) Rõ ràng $10^{11}-1< 10^{12}-1$. 

Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$

Áp dụng kết quả phần a:

$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$

Lời giải:

$B=\frac{10^{11}+10}{10^{12}+10}$

Đặt $10^{11}-1=a; 10^{12}-1=b$ thì $0< a< b$. Khi đó:

$A-B=\frac{a}{b}-\frac{a+11}{b+11}=\frac{11(a-b)}{b(b+11)}<0$

$\Rightarrow A< B$

Dễ vãi

a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)

c) \(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=\left(7^3\right)^{100}=343^{100}>243^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

Giải chi tiết giúp mình ạ~

\(\left(d\right):202^{303}=\left(202^3\right)^{101}=8242408^{101}>303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(\left(e\right):107^{50}=\left(107^2\right)^{25}=11449^{25}< 73^{75}=\left(73^3\right)^{25}=389017^{25}\)

Lời giải:
a.

\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)

\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)

b.

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

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

$\Rightarrow 10A< 10B\Rightarrow A< B$

2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

A=-2015/2015x2016

A=-1/2016

B=-2014/2014x2015

B=-1/2015

vi 2016>2015,-1/2016>-1/2015

vay A>B

b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)

\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)

Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)

\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)

Ta có: \(10^{2010}+1< 10^{2011}+1\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)

\(\Leftrightarrow10A>10B\)

hay A>B

Giải:

Ta có:

A=20¹⁰+1/20¹⁰-1

A=20¹⁰-1+2/20¹⁰-1

A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-1<2/20¹⁰-3 nên A<B

Chúc bạn học tốt!

Lời giải:
$10A=\frac{10^{2021}-10}{10^{2021}-1}=\frac{10^{2021}-1-9}{10^{2021}-1}$

$=1-\frac{9}{10^{2021}-1}>1$

$10B=\frac{10^{2022}+10}{10^{2022}+1}=\frac{10^{2022}+1+9}{10^{2022}+1}$

$=1+\frac{9}{10^{2022}+1}<1$

$\Rightarrow 10A> 1> 10B$

Suy ra $A> B$