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Bài 2:
Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)
\(37^{1320}=37^{2\cdot660}=1369^{660}\)
mà \(1331^{660}< 1369^{660}\)
nên \(11^{1979}< 37^{1320}\)
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, A = 3020 xx 3110 - 5 = 3020 xx 3109 + 3020 - 5`
`= 3020 xx 3109 + 3015 = B`.
`b, B = (2022-2)(2022+2) = 2022^2-4 < 2022^2 = A.`
bài 1 :
vì x chia hết cho 10 và 22 nên x là BC của 10 và 22 mà x<120
ta có : Ư(10) = { 1;2;5;10 }
Ư(22) = { 1;2;11;22 }
ƯC(10;22) = { 1;2 }
suy ra : x thuộc ( kí hiệu thuộc ) { 1;2 }
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
(x-1)2020=(x-1)2022
=>(x-1)2020-(x-1)2022=0
=>(x-1)2020-(x-1)2020.(x-1)2=0
=>(x-1)2020(1-(x-1)2=0
=>(x-1)2020=0 hoặc 1-(x-1)2=0
=>x=1 hoặc x=2.
Bài 2
a,2105 và 545
2105=(27)15=12815
545=(53)15=12515
Vì 12815>12515 nên 2105>545.
b,
554 và 381
554=(56)9=156259
381=(39)9=196839
Vì 156259<196839 nên 554<381
Bài 1 :
\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)
\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)
\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)