So sánh:
A= 17 mũ 18 + 1 trên 17 mũa 19 +1
B= 17 mũ 17 + 1 trên 17 mũ 18 + 1
giúp tớ nha
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Ta có:
\(A=\frac{17^{18}+1}{17^{19}+1}\)
\(\Rightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)
\(\Rightarrow17A=1+\frac{16}{17^{19}+1}\)
\(B=\frac{17^{17}+1}{17^{18}+1}\)
\(\Rightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)
\(\Rightarrow17B=1+\frac{16}{17^{18}+1}\)
Vì \(\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\Rightarrow17A< 17B\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
\(A=\dfrac{10^{17}+3}{10^{17}+1}=1+\dfrac{2}{10^{17}+1}\\ B=\dfrac{10^{18}+1}{10^{18}-1}=1+\dfrac{2}{10^{18}-1}=1+\dfrac{2}{10^{17}+1+\left(9\cdot10^{17}-2\right)}\)
Ta có : \(9\cdot10^{17}-2>0\Rightarrow10^{17}+1+\left(9\cdot10^{17}-2\right)>10^{17}+1\\ \Rightarrow\dfrac{2}{10^{17}+1}>\dfrac{2}{10^{18}-1}\Rightarrow A>B\)
#)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
B = 2^20 - ( 2^19 + 2^18 + 2^17 + ... + 2^1 + 2^0
B = (2^19 x 2 - 2^19) - 2^18 - 2^17 - ... - 2^1 - 0
B = (2^18 x 2 - 2^18) - 2^17 - 2^16 - ... - 2^1 - 0
B = (2^17 x 2 - 2^17) - 2^16 - 2^15 - ... - 2^1 - 0
Cứ tách xong lại đóng ngoặc cho đến:
B = 2^1 x 2 - 2^1 - 0
B = 2^1 - 0
B = 2
Đặt A=219+218+...+20
2A=2(219+218+...+1)
2A=220+219+...+2
2A-A=(220+219+...+2)-(219+218+...+2)
A=220-2
Thay A vào B ta có: 220-(220-2)
=220-220+2
=0+2=2
Ta có: 2B = \(2^{21}-2^{20}-...-2^0\)
B = \(2^{20}-2^{19}-...-2^1\)
=> 2B + B = \(2^{21}-1\)
=> 3B = \(2^{21}-1\)
=> B = \(\frac{2^{21}-1}{3}\)
2^20 - 2^19 - 2^18 - 2^17 - ... - 2^1 - 2^0
=(2^19 x 2 - 2^19) - 2^18 - 2^17 - 2^16 - ... - 2^1
=(2^18 x 2 - 2^18) - 2^17 - 2^16 - 2^15 - ... - 2^1
Cứ tính xong trong ngoặc lại ghép tiếp........
=2^1 x 2 - 2^1
=2^1
=2
Sửa đề: \(C=\dfrac{17^{99}+1}{17^{99}-1}\)
\(C=\dfrac{17^{99}-1+2}{17^{99}-1}=1+\dfrac{2}{17^{99}-1}\)
\(D=\dfrac{17^{98}-1+2}{17^{98}-1}=1+\dfrac{2}{17^{98}-1}\)
17^99>17^98
=>17^99-1>17^98-1
=>C<D
Ta có:
A=1718+11719+1A=1718+11719+1
⇒17A=1719+1+161719+1⇒17A=1719+1+161719+1
⇒17A=1+161719+1⇒17A=1+161719+1
B=1717+11718+1B=1717+11718+1
⇒17B=1718+1+161718+1⇒17B=1718+1+161718+1
⇒17B=1+161718+1⇒17B=1+161718+1
Vì 161719+1<161718+1⇒17A<17B161719+1<161718+1⇒17A<17B
⇒A<B⇒A<B
Vậy A<B