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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?
#)Giải :
\(A=\frac{20^{18}+1}{20^{19}+1}\)và \(B=\frac{20^{17}+1}{20^{18}+1}\)
\(A=\frac{20^{18}+1}{20^{18+1}+1}\)và \(B=\frac{20^{17}+1}{20^{17+1}+1}\)
\(A=\frac{1}{20+1}\)và \(B=\frac{1}{20+1}\)
\(A=\frac{1}{21}\)và \(B=\frac{1}{21}\)
\(\Rightarrow A=B\)
#~Will~be~Pens~#
A>2018 +1+19/2019 +1+19
A>2018+20/2019+20
A>20(2017+1)/20(2018+1)
A>2017+1/2018+1
=>A>B
Chúc bạn học tốt
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
\(A=\dfrac{113^{20}+113-112}{113^{19}+1}=113-\dfrac{112}{113^{19}+1}\)
\(B=\dfrac{113^{19}+113-112}{113^{18}+1}=113-\dfrac{112}{113^{18}+1}\)
mà \(113^{19}+1>113^{18}+1\)
nên \(A>B\)
áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)
Ta có: \(A=\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}\)\(=\frac{17^{18}-17}{17^{20}-17}=\frac{17.\left(17^{17}-1\right)}{17.\left(17^{19}-1\right)}\)\(=\frac{17^{17}-1}{17^{19}-1}\)
\(\Rightarrow A< B\)
\(A=\frac{17^{18}-1}{17^{20}-1}\Rightarrow17^2A=\frac{17^{18}-1}{17^{18}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}\left(1\right)\)
\(B=\frac{17^{17}-1}{17^{19}-1}\Rightarrow17^2B=\frac{17^{17}-1}{17^{17}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(2\right)\)
\(\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}< \frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\Rightarrow1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}>1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(3\right)\)
Từ \(\left(1\right);\left(2\right)\&\left(3\right)\Rightarrow17^2A>17^2B\Leftrightarrow A>B.\)
\(\dfrac{1}{13}A=\dfrac{13^{19}+1}{13^{19}+\dfrac{1}{13}}=1+\dfrac{\dfrac{12}{13}}{13^{19}+\dfrac{1}{13}}\)
\(\dfrac{1}{13}B=\dfrac{13^{20}+1}{13^{20}+\dfrac{1}{13}}=1+\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}\)
Vì \(\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}< \dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}\Rightarrow1+\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}< 1+\dfrac{\dfrac{12}{13}}{13^{20}+\dfrac{1}{13}}\)
\(\Rightarrow\dfrac{1}{13}A>\dfrac{1}{13}B\Rightarrow A>B\)
Vậy...
Ta xét hiệu:
\(A-1=\dfrac{3^{19}+1}{3^{18}+1}-1=\dfrac{3^{19}-3^{18}}{3^{18}+1}=\dfrac{3^{18}.2}{3^{18}+1}\)
\(B-1=\dfrac{3^{20}+1}{3^{19}+1}-1=\dfrac{3^{20}-3^{19}}{3^{19}+1}=\dfrac{3^{19}.2}{3^{19}+1}\)
Xét: \(\dfrac{A-1}{B-1}=\dfrac{3^{18}.2}{3^{18}+1}\cdot\dfrac{3^{19}+1}{3^{19}.2}=\dfrac{3^{19}+1}{\left(3^{18}+1\right).3}=\dfrac{3^{19}+1}{3^{19}+3}< 1\)
=> A-1<B-1
=>A<B
Ta có: \(\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}=\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+1\)
\(=\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}=20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)\)
Thế lại bài toán ta được
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=\dfrac{20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)
Ta có
\(\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}\\ =\dfrac{1}{19}+1+\dfrac{2}{18}+1+\dfrac{3}{17}+1+...+\dfrac{19}{1}+1-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{1}-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+20-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{2}+1+19-19\\ =\dfrac{20}{20}+\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}\\ =20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)\)
Thế vào ta có:
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\\ =\dfrac{20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)}{\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}}\\ =20\)
\(=\dfrac{\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+\left(\dfrac{18}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}\)
\(=\dfrac{\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)
A<B vì các số mũ đều giống nhau nên sẽ s2 phần số nguyên vậy nên nếu s2 5 với 6 thì 5 bé hơn nên A<B
Ta có: \(20A=\dfrac{20^{19}+20}{20^{19}+1}=1+\dfrac{19}{20^{19}+1}\)
\(20B=\dfrac{20^{18}+20}{20^{18}+1}=1+\dfrac{19}{20^{18}+1}\)
Vì \(\dfrac{19}{20^{19}+1}< \dfrac{19}{20^{18}+1}\Rightarrow1+\dfrac{19}{20^{19}+1}< 1+\dfrac{19}{20^{18}+1}\)
\(\Rightarrow20A< 20B\Rightarrow A< B\)
Vậy A < B
Ta có: \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)(a \(\in\) N và b,c,d \(\in\) N*)
Áp dụng kiến thức đó, ta được:
A = \(\dfrac{20^{18}+1}{20^{19}+1}\) <\(\dfrac{20^{18}+1+19}{20^{19}+1+19}\)= \(\dfrac{20^{18}+20}{20^{19}+20}\) = \(\dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}\)
= \(\dfrac{20^{17}+1}{20^{18}+1}\) = B
Vậy A < B