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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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a,
15^12=(3*5)^12=3^12*5^12
81^3*125^5=(3^4)^3*(5^3)^5=3^12*5^15
Vì 12<15 suy ra 5^12<5^15
Suy ra 3^12*5^12<3^12*5^15
\(a.81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}=3^{12}.5^{12}.5^3=\left(3.5\right)^{12}.5^3=15^{12}.5^3>15^{12}\)
\(b.4^{20}.81^{12}=\left(2^2\right)^{20}.\left(9^2\right)^{12}=2^{40}.9^{24}=2^{20}.2^{20}.9^{20}.9^4=\left(2.9\right)^{20}.2^{20}.9^4=18^{20}.2^{20}.9^4>18^{20}\)
\(c.73^{75}=\left(73^3\right)^{25}=389017^{25}\)
\(107^{50}=107^{2.50}=\left(107^2\right)^{25}=11449^{25}\)
Vì \(389017^{25}>11449^{25}\Rightarrow73^{75}>107^{50}\)
1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)
=(1-1/3)....0.....(1-9/5)
=0
=>đpcm.
b)ta xét:
1/22 = 1/2x2 < 1/1x2
.............
1/82 = 1/8x8 <1/7x8
=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8
<=> B <1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8
<=> B < 1 - 1/8 = 7/8 < 1
=> B < 1 => đpcm
2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)
Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)
Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)
=> A > B
b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C
=> C > D
c)gọi 2010 là a
ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)
áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)
=> E > F
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
A = 1 + 2 + 22 + ... + 220
2A = 2 + 22 + 23 + ... + 221
2A - A = (2 + 22 + 23 + ... + 221) - (1 + 2 + 22 + ... + 220)
A = 221 - 1 < 221 = B
=> A < B
A = 1 + 2 + 22
+ ... + 220
2A = 2 + 22
+ 23
+ ... + 221
2A - A = (2 + 22
+ 23
+ ... + 221) - (1 + 2 + 22
+ ... + 220)
A = 221
- 1 < 221
= B
=> A < B
k cho mk nha $_$
:D
#)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
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