Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{10^{17}+3}{10^{17}+1}=1+\dfrac{2}{10^{17}+1}\\ B=\dfrac{10^{18}+1}{10^{18}-1}=1+\dfrac{2}{10^{18}-1}=1+\dfrac{2}{10^{17}+1+\left(9\cdot10^{17}-2\right)}\)
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
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
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 \(A< B\)
cho mk hỏi là tại sao lại là 17.A?