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Bài 1:
1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà \(17^{19}+1>17^{18}+1\)
nên 17A>17B
hay A>B
2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)
\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)
mà \(98^{89}+1>98^{88}+1\)
nên C>D
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?
Bài này có rất nhiều cách lm nhé!
Ta có : A = \(\dfrac{17^{18}+1}{17^{19}+1}\) => 17A = \(\dfrac{17^{19}+17}{17^{19}+1}\) = \(1+\dfrac{16}{17^{19}+1}\)
B = \(\dfrac{17^{17}+1}{17^{18}+1}\) => 17B = \(\dfrac{17^{18}+17}{17^{18}+1}\) = \(1+\dfrac{16}{17^{18}+1}\)
Vì \(\dfrac{16}{17^{19}+1}\) < \(\dfrac{16}{17^{18}+1}\) ( vì 1719 +1 > 1716+1 )
=> \(1+\dfrac{16}{17^{19}+1}\) < \(1+\dfrac{16}{17^{18}+1}\)
=> 17A < 17B
=> A < B ( vì 17 > 0)
Ta có :
\(A=\dfrac{17^{18}+1}{17^{19}+1}\)
17A= \(17\times\dfrac{17^{18}+1}{17^{19}+1}\)
\(17A=\dfrac{17^{19}+17}{17^{19}+1}\)
\(17A=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}\)
\(17A=\dfrac{17^{19}+1}{17^{19}+1}+\dfrac{16}{17^{19}+1}\)
\(17A=1+\dfrac{16}{17^{19}+1}\)
Lại có :
\(B=\dfrac{17^{17}+1}{17^{18}+1}\)
\(17B=17\times\dfrac{17^{17}+1}{17^{18}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}\)
\(17B=\dfrac{\left(17^{18}+1\right)+16}{17^{18}+1}\)
\(17B=\dfrac{17^{18}+1}{17^{18}+1}+\dfrac{16}{17^{18}+1}\)
\(17B=1+\dfrac{16}{17^{18}+1}\)
Mà : \(\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)
\(\Rightarrow1+\dfrac{16}{17^{19}+1}< 1+\dfrac{16}{17^{18}+1}\)
⇒ A < B
Vậy A < B
a, \(\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)
b, \(-\dfrac{24}{17}+\dfrac{7}{17}+\dfrac{1}{16}=\dfrac{-17}{17}+\dfrac{1}{16}=-1+\dfrac{1}{16}=-\dfrac{15}{16}\)
a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà 17^19+1>17^18+1
nên A<B
b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
2^2021-1<2^2022-1
=>1/2^2021-1>1/2^2022-1
=>-1/2^2021-1<-1/2^2022-1
=>C<D
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)
B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)
vậy A=B