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\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
2023>2022
=>10^2023+1>10^2022+1
=>10A<10B
=>A<B
M=\(\dfrac{10^{100^{ }}+1}{10^{101}+1}\)
M=\(\dfrac{10^{99+1}+1}{10^{100+1}+1}\)
M=\(\dfrac{10^{99}.10+1}{10^{100}.10+1}\)
N=\(\dfrac{10^{99^{ }}+1}{10^{100}+1}\)
=>M lớn hơn N
M>N,vì:\(\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{100}}{10^{101}}\)
\(\dfrac{10^{99}+1}{10^{100}+1}=\dfrac{10^{99}}{10^{100}}\)
\(\dfrac{10^{100}}{10^{101}}>\dfrac{10^{99}}{10^{100}}\)
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)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 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
17A=1719+1+16/1719+1
17A=1+16/1719+1
phần in nghiêng mình không hiểu lắm, bn giải thích cho mình được ko?
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ < \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{97}{98}.\dfrac{98}{99}< \dfrac{1}{99}\\ < \dfrac{1}{10}.\\\\ =>A< \dfrac{1}{10}\)
ta có:
1/10.A=10100+1/10(1099+1)
1/10.A=10100+1/10100+10
1/10.A=1-(9/10100+10)
1/10.B=10101+1/10(10100+1)
1/10.B=10101+1/10101+10
1/10.B=1-(9/10101+10)
vì(10101+10)>(10100+1)=> 9/10101+10 < 9/10100+10 => 1-(9/10101+10) > 1-(9/10100+10)
hay 1/10.A>1/10.B
=>A>B
ta có:
1/10.A=10100+1/10(1099+1)
1/10.A=10100+1/10100+10
1/10.A=1-(9/10100+10)
1/10.B=10101+1/10(10100+1)
1/10.B=10101+1/10101+10
1/10.B=1-(9/10101+10)
vì(10101+10)>(10100+1)=> 9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)
hay 1/10.A<1/10.B
=>A<B