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a: Ta có: \(-\left(x+5\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x+5\right)^2+2021\le2021\forall x\)
Dấu '=' xảy ra khi x=-5
A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
Minh tinh ra duoc la:
a=2020^99-2/2019
2020^99-1/2019 2020^99-2
2020^99-1 2020^99*2019-2*2019
2*2019-1 2020^99(2019-1)
2*2019-2 2020^99*2018
Gio phai lam sao tiep moi nguoi?
Bài 2:
Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)
\(37^{1320}=37^{2\cdot660}=1369^{660}\)
mà \(1331^{660}< 1369^{660}\)
nên \(11^{1979}< 37^{1320}\)
Ta có: \(\frac{A}{2020}=\frac{2020^{100}-1}{2020^{100}+2020}=\frac{2020^{100}+2020-2021}{2020^{100}+2020}=1-\frac{2021}{2020^{100}+2020}\)
\(\frac{B}{2020}=\frac{2020^{101}-1}{2020^{101}+2020}=\frac{2020^{101}+2020-2021}{2020^{101}+2020}=1-\frac{2021}{2020^{101}+2020}\)
Ta có: \(2020^{100}+2020<2020^{101}+2020\)
=>\(\frac{2021}{2020^{100}+2020}>\frac{2021}{2020^{101}+2020}\)
=>\(-\frac{2021}{2020^{100}+2020}<-\frac{2021}{2020^{101}+2020}\)
=>\(-\frac{2021}{2020^{100}+2020}+1<-\frac{2021}{2020^{101}+2020}+1\)
=>\(\frac{A}{2020}<\frac{B}{2020}\)
=>A<B