Câu 1.99²⁰và 9999¹⁰

⁼(99²)¹⁰=9801¹⁰

Vậy 9801¹⁰<9999¹⁰ suy ra  99²⁰<9999¹⁰

Câu 2. 3⁵⁰⁰và 7³⁰⁰

 3⁵⁰⁰=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=(7³)¹⁰⁰=343¹⁰⁰

Vậy 243¹⁰⁰<343¹⁰⁰ => 3⁵⁰⁰<7³⁰⁰

f: 11^1979<11^1980=1331^660

37^1320=(37^2)^660=1369^660

1331<1369

=>1331^660<1369^660

=>11^1980<37^1320

=>11^1979<37^1320

g: 10^10=2^10*5^10

48*50^5=2^4*3*2^5*5^10=2^9*3*5^10

2^10<2^9*3

=>2^10*5^10<2^9*3*5^10

=>10^10<48*50^5

Đáp án là :

a) <

b, <

c, chưa biết

d, <

đúng thì ủng hộ tớ nha

a)3⁵⁰⁰ = (3⁵)¹⁰⁰ = 243¹⁰⁰

7³⁰⁰ = (7³)¹⁰⁰ = 343¹⁰⁰

243¹⁰⁰ < 343¹⁰⁰ => 3⁵⁰⁰ < 7³⁰⁰

a) \(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)

\(9999^{10}=\left(99.101\right)^{10}\)

Ta thấy \(99.100>99.99\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\Leftrightarrow99^{20}< 9999^{10}\)

b) Ta có : \(202^{303}=\left[\left(2.101\right)^3\right]^{101}=8^{101}.101^{303}\)

\(303^{202}=\left[\left(3.101\right)^2\right]^{101}=9^{101}.101^{202}\)

Tự làm tiếp nha bn

a)99²⁰ và 999¹⁰

Ta có:ƯCLN(20;10)=10

\(\Rightarrow99^{20}=\left(99^2\right)^{10}\)

\(9999^{10}=\left(9999^1\right)^{10}\)

\(99^2=9801< 9999\)

\(\Rightarrow99^{20}< 9999^{10}\)

câu 1 >

câu 2 <

câu 3 >

câu 4 >

caua <

\(202^{303}=\left(101.2\right)^{303}=101^{606}\)

\(303^{202}=\left(101.3\right)^{202}=101^{606}\)

Vì 101⁶⁰⁶ = 101⁶⁰⁶ nên 202³⁰³ = 303²⁰²

a, 202²⁰³=(101.2)²⁰³

=101²⁰³.2²⁰³

=101²⁰².2²⁰².202

b, 203²⁰²=(101,5.2)²⁰²

=101,5²⁰².2²⁰²

còn lại dễ

b, 1990¹⁰+1990⁹=1990⁹.1990+1990⁹=1990⁹.(1990+1)=1990⁹.1991

1991²⁰=1991¹⁹.1991

=>1990¹⁰+1990⁹<1991²⁰

c, 11¹⁹⁷⁹<11¹⁹⁸⁰=(11³)⁶⁶⁰=1331⁶⁶⁰

37¹³²⁰=(37²)⁶⁶⁰=1369⁶⁶⁰

=>11¹⁹⁷⁹<37¹³²⁰

Lời giải:
$1990^{10}+1990^9=1990^9(1990+1)=1991.1990^9< 1991.1991^9=1991^{10}$

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$10^{10}=(10^2)^5=100^5=(2.50)^5=2^5.50^5=32.50^5< 48.50^5$

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$11^{1979}< 11^{1980}=(11^3)^{660}=1331^{660}$
$37^{1320}=(37^2)^{660}=1369^{660}> 1331^{660}$

$\Rightarrow 11^{1979}< 37^{1320}$